Solucao_Calculo_5ed_SEC 7 1a7 4

Prévia do material em texto

u=ln x dv=xdx�du=dx/x v= 12 x
2 �udv=uv��vdu
�xln xdx = 12 x
2
ln x�� 12 x
2 dx/x( )= 12 x
2
ln x�
1
2 �xdx=
1
2 x
2
ln x�
1
2 �
1
2 x
2
+C
= 12 x
2
ln x�
1
4 x
2
+C
u=� dv=sec 2� d� �du=d� v=tan�
�� sec 2� d� =� tan� �� tan� d� =� tan� �ln sec� +C
u=x dv=cos 5xdx�du=dx v= 15 sin 5x
�xcos 5xdx= 15 xsin 5x��
1
5 sin 5xdx=
1
5 xsin 5x+
1
25 cos 5x+C
u=x dv=e�xdx�du=dx v=�e�x �xe�xdx=�xe�x+�e�xdx=�xe�x�e�x+C
u=r dv=er/2dr�du=dr v=2er/2 �rer/2dr=2rer/2��2er/2dr=2rer/2�4er/2+C
u=t dv=sin 2t dt�du=dt v=� 12 cos 2t �tsin 2t dt=�
1
2 tcos 2t+
1
2 �cos 2t
dt=� 12 tcos 2t+
1
4 sin 2t+C
u=x
2 dv=sin� xdx�du=2xdx v=� 1� cos� x
I=�x2sin� xdx=� 1� x
2
cos� x+
2
� �xcos� xdx *
U =x dV =cos� xdx�dU =dx V = 1� sin� x
�xcos� xdx= 1� xsin� x�
1
� �sin� xdx=
1
� xsin� x+
1
�
2
cos� x+C
1
�xcos� xdx
*
I=�
1
� x
2
cos� x+
2
�
1
� xsin� x+
1
�
2
cos� x+C
1
=�
1
� x
2
cos� x+ 2
�
2
xsin� x+ 2
�
3
cos� x+C
C=
2
� C1
1. Let , , . Then by Equation 2, ,
2. Let , , . Then
 .
3. Let , , . Then by Equation 2,
 .
4. Let , , . Then .
5. Let , , . Then .
6. Let , , . Then 
 .
7. Let , and . Then
 ( ).
Next let , , , so
 . Substituting for in (
 ), we get
 ,
where .
 1
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
u=x
2 dv=cos mxdx�du=2xdx v= 1
m
sin mx
I=�x2cos mxdx= 1
m
x
2
sin mx�
2
m
�xsin mxdx * U =x dV =sin mxdx�dU =dx
V =�
1
m
cos mx �xsin mxdx=� 1
m
xcos mx+
1
m
�cos mxdx=� 1
m
xcos mx+ 1
m
2
sin mx+C
1
�xsin mxdx *
I=
1
m
x
2
sin mx�
2
m
�
1
m
xcos mx+ 1
m
2
sin mx+C
1
=
1
m
x
2
sin mx+ 2
m
2
xcos mx� 2
m
3
sin mx+C
C=�
2
m
C
1
u=ln (2x+1) dv=dx�du= 22x+1 dx v=x
� ln (2x+1)dx =xln (2x+1)�� 2x2x+1 dx=xln (2x+1)��
(2x+1)�1
2x+1 dx
=xln (2x+1)�� 1� 12x+1 dx=xln (2x+1)�x+
1
2 ln (2x+1)+C
=
1
2 (2x+1)ln (2x+1)�x+C
u=sin
�1
x dv=dx�du= dx
1�x
2
v=x �sin �1xdx=xsin �1x�� x
1�x
2
dx
t=1�x
2 dt=�2xdx �� xdx
1�x
2
=��t�1/2 � 12 dt =
1
2 2t
1/2( )+C=t1/2+C= 1�x2 +C
�sin �1xdx=xsin �1x+ 1�x2 +C
u=arctan 4t dv=dt�du= 4
1+(4t)
2
dt= 4
1+16t
2
dt v=t
� arctan 4t dt =t arctan 4t��
4t
1+16t
2
dt=t arctan 4t� 18 �
32t
1+16t
2
dt
=t arctan 4t�
1
8 ln (1+16t
2
)+C
u=ln p dv=p5dp�
8. Let , , . Then
 ( ). Next let , ,
 , so .
Substituting for in ( ), we get
 ,
where .
9. Let , , . Then
10. Let , , . Then . Setting
 , we get , so .
Hence, .
11. Let , , . Then
12. Let , 
 2
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
du= 1p dp v=
1
6 p
6 � p5ln pdp= 16 p
6
ln p�
1
6 � p
5dp= 16 p
6
ln p�
1
36 p
6
+C
u= ln x( ) 2 dv=dx�du=2ln x� 1
x
dx v=x
I=�(ln x)2dx=x(ln x)2�2�xln x� 1
x
dx=x(ln x)2�2� ln xdx U =ln x dV =dx�dU =1/xdx V =x
� ln xdx=xln x��x� 1/x( ) dx=xln x��dx=xln x�x+C1
I=x(ln x)
2
�2 xln x�x+C
1( )=x(ln x)2�2xln x+2x+C C=�2C1
u=t
3 dv=et dt�du=3t2dt v=et I=�t3et dt=t3et��3t2et dt
dv=et dt
I = t
3
e
t
� 3t
2
e
t
��6tet dt( )=t3et�3t2et+6tet��6et dt
= t
3
e
t
�3t
2
e
t
+6te
t
�6e
t
+C= t3�3t2+6t�6( ) et+C
p(t) n t
� p t( ) et dt= p(t)�p /(t)+p / /(t)�p / / /(t)+ � � � + �1( ) np n( ) (t) et+C
u=sin 3� dv=e2� d��du=3cos 3� d� v= 12 e
2�
I=�e2�sin 3� d� = 12 e
2�
sin 3� �
3
2 �e
2�
cos 3� d� U =cos 3� dV =e2� d�
� dU =�3sin 3� d� V = 12 e
2�
�e2� cos 3� d� = 12 e
2�
cos 3� +
3
2 �e
2�
sin 3� d�
I=
1
2 e
2�
sin 3� �
3
4 e
2�
cos 3� �
9
4 �e
2�
sin 3� d� = 12 e
2�
sin 3� �
3
4 e
2�
cos 3� �
9
4 I�
13
4 I=
1
2 e
2�
sin 3� �
3
4 e
2�
cos 3� +C
1
I=
1
13 e
2�
2sin 3� �3cos 3�( )+C C= 413 C1
u=e
�� dv=cos 2� d��du=�e�� d� v= 12 sin 2�
I=�e�� cos 2� d� = 12 e
��
sin 2� �� 12 sin 2� �e
�� d�( )= 12 e
��
sin 2� +
1
2 �e
��
sin 2� d�
U =e
�� dV =sin 2� d��dU =�e�� d� V =� 12 cos 2�
 , . Then .
13. First let , , . Then by Equation 2,
 . Next let , , 
to get . Thus,
 , where .
14. Let , , . Then . Integrate by parts twice
more with .
More generally, if is a polynomial of degree in , then repeated integration by parts shows that
 .
15. First let , , . Then
 . Next let , 
 , to get
 . Substituting in the previous formula gives
 . Hence, , where .
16. First let , , . Then
 .
Next let , , , so
 3
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�e��sin 2� d� =� 12 e
��
cos 2� �� � 12 cos 2� �e
�� d�( )=� 12 e
��
cos 2� �
1
2 �e
��
cos 2� d�
I=
1
2 e
��
sin 2� +
1
2 �
1
2 e
��
cos 2� �
1
2 I =
1
2 e
��
sin 2� �
1
4 e
��
cos 2� �
1
4 I�
5
4 I=
1
2 e
��
sin 2� �
1
4 e
��
cos 2� +C
1
�
I=
4
5
1
2 e
��
sin 2� �
1
4 e
��
cos 2� +C
1
=
2
5 e
��
sin 2� �
1
5 e
��
cos 2� +C
u=y dv=sinh ydy�du=dy v=cosh y
� ysinh ydy=ycosh y��cosh ydy=ycosh y�sinh y+C
u=y dv=cosh aydy�du=dy v= sinh ay
a
� ycosh aydy= ysinh ay
a
�
1
a
�sinh aydy= ysinh ay
a
� cosh ay
a
2
+C
u=t dv=sin 3t dt�du=dt v=� 13 cos 3t
��
0
t sin 3t dt= � 13 t cos 3t
�
0
+
1
3 �
�
0
cos 3t dt= 13 � �0 +
1
9 sin 3t
�
0
=
�
3 .
u=x
2
+1 dv=e�xdx�du=2xdx v=�e�x
�1
0
x
2
+1( ) e�xdx= � x2+1( ) e�x 1
0
+�1
0
2xe
�xdx=�2e�1+1+2�1
0
xe
�xdx U =x dV =e�xdx�dU =dx
V =�e
�x �1
0
xe
�xdx= �xe�x
1
0
+�1
0
e
�xdx=�e�1+ �e�x
1
0
=�e
�1
�e
�1
+1=�2e
�1
+1
�1
0
x
2
+1( ) e�xdx=�2e�1+1+2 �2e�1+1( )=�2e�1+1�4e�1+2=�6e�1+3
u=ln x dv=x�2dx�du= 1
x
dx v=�x�1
�
1
2
ln x
x
2
dx= � ln x
x
2
1
+�
1
2
x
�2dx=� 12 ln 2+ln 1+ �
1
x
2
1
=�
1
2 ln 2+0�
1
2 +1=
1
2 �
1
2 ln 2
u=ln t dv= t dt�du=dt/t v= 23 t
3/2
 .
So
 .
17. Let , , . Then
 .
18. Let , , . Then
 .
19. Let , , . Then
20. First let , , . By (6),
 . Next let , 
, . By (6) again, . So
 .
21. Let , , . By (6),
 .
22. Let , , . By Formula 6,
 4
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�4
1
t ln t dt= 23 t
3/2
ln t
4
1
�
2
3 �
4
1
t dt= 23 � 8� ln 4�0�
2
3 �
2
3 t
3/2 4
1
=
16
3 ln 4�
4
9 8�1( )=
16
3 ln 4�
28
9
u=y dv= dy
e
2y
=e
�2ydy�du=dy v=� 12 e
�2y
�1
0
y
e
2y
dy= � 12 ye
�2y 1
0
+
1
2 �
1
0
e
�2ydy= � 12 e
�2
+0 �
1
4 e
�2y 1
0
=�
1
2 e
�2
�
1
4 e
�2
+
1
4 =
1
4 �
3
4 e
�2
u=x dv=csc 2xdx�du=dx v=�cot x
�� /2
� /4
xcsc
2
xdx = �xcot x � /2
� /4
+�� /2
� /4
cot xdx=� �2 � 0+
�
4 � 1+ ln sin x
� /2
� /4
=
�
4 +ln 1�ln
1
2
=
�
4 +0�ln 2
�1/2
=
�
4 +
1
2 ln 2
u=cos
�1
x dv=dx�du=� dx
1�x
2
v=x
I= �
0
1/2
cos
�1
xdx= xcos �1x
1/2
0
+�
0
1/2
xdx
1�x
2
=
1
2 �
�
3 + �1
3/4
t
�1/2
�
1
2 dt t=1�x
2
� dt=�2xdx
I=
�
6 +
1
2 �3/4
1
t
�1/2dt= �6 + t
1
3/4
=
�
6 +1�
3
2 =
1
6 � +6�3 3( )
u=x dv=5xdx�du=dx v= 5x /ln 5( )
�
0
1
x5
xdx =
x5
x
ln 5
1
0
��
0
1
5
x
ln 5 dx=
5
ln 5 �0�
1
ln 5
5
x
ln 5
1
0
=
5
ln 5 �
5
ln 5( ) 2
+ 1
ln 5( ) 2
=
5
ln 5 �
4
ln 5() 2
u= ln sin x( ) dv=cos xdx�du= cos xsin x dx v=sin x
I=�cos x ln sin x( ) dx=sin x ln sin x( )��cos xdx=sin x ln sin x( )�sin x+C
.
23. Let , , . Then
 .
24. Let , , . Then
 [see Exercise 5.5.]
25. Let , , . Then
 , where .
Thus, .
26. Let , , . Then
27. Let , , . Then
 .
 5
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
t=sin x dt=cos xdx I=� ln t dt=t ln t�t+C
I=sin x ln sin x�1( )+C
u=arctan(1/x) dv=dx�du= 1
1+(1/x)
2
� �1
x
2
dx= �dx
x
2
+1
v=x
�
1
3
arctan(1/x)dx = xarctan 1
x
3
1
+ �
1
3
x
dx x
2
+1= 3
�
6 �1�
�
4 +
1
2 ln (x
2
+1)
3
1
=
� 3
6 �
�
4 +
1
2 (ln 4�ln 2)=
� 3
6 �
�
2 +
1
2 ln
4
2 =
� 3
6 �
�
2 +
1
2 ln 2
w=ln x�dw=dx/x x=ew dx=ewdw
�cos ln x( ) dx = �ewcos wdw= 12 e
w
sin w+cos w( )+C
= 12 x sin ln x( )+cos ln x( ) +C
u=r
2 dv= r
4+r
2
dr�du=2r dr v= 4+r2
�
0
1
r
3
4+r
2
dr
= r
2
4+r
2 1
0
�2�
0
1
r 4+r
2 dr= 5� 23 4+r
2( ) 3/2 1
0
= 5�
2
3 (5)
3/2
+
2
3 (8)= 5 1�
10
3 +
16
3 =
16
3 �
7
3 5
u=(ln x)
2 dv=x4dx�du=2 ln x
x
dx v= x
5
5
�
1
2
x
4
(ln x)
2dx= x
5
5 ln x( )
2
2
1
�2�
1
2
x
4
5 ln xdx=
32
5 ln 2( )
2
�0�2�
1
2
x
4
5 ln xdx
U =ln x dV = x
4
5 dx�dU =
1
x
dx V = x
5
25
�
1
2
x
4
5 ln xdx=
x
5
25 ln x
2
1
��
1
2
x
4
25 dx=
32
25 ln 2�0�
x
5
125
2
1
=
32
25 ln 2�
32
125 �
1
125
Another method: Substitute , so . Then (see Example 2) and
so .
28. Let , , . Then
29. Let . Then and , so
 [by the method of Example 4]
30. Let , , . By (6),
31. Let , , . By (6),
 .
Let , , .
Then .
 6
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�
1
2
x
4
(ln x)
2dx= 325 (ln 2)
2
�2
32
25 ln 2�
31
125 =
32
5 (ln 2)
2
�
64
25 ln 2+
62
125
u=sin (t�s) dv=esds�du=�cos (t�s)ds v=es
I=�t
0
e
s
sin (t�s)ds= essin (t�s)
t
0
+�t
0
e
s
cos (t�s)ds=etsin 0�e0sin t+I
1
I
1
U =cos (t�s) dV =esds
�dU =sin (t�s)ds V =es I
1
= e
s
cos (t�s)
t
0
��t
0
e
s
sin (t�s)ds=etcos 0�e0cos t�I
I=�sin t+e
t
�cos t�I�2I=e
t
�cos t�sin t� I=
1
2 e
t
�cos t�sin t( )
w= x x=w
2 dx=2wdw �sin x dx=�2wsin wdw
u=2w dv=sin wdw du=2dw v=�cos w
�2wsin wdw = �2wcos w+�2cos wdw=�2wcos w+2sin w+C
= �2 x cos x +2sin x +C=2 sin x � x cos x( )+C
w= x x=w
2 dx=2wdw �4
1
e
x
dx=�2
1
e
w
2wdw u=2w
dv=ewdw du=2dw v=ew �2
1
e
w
2wdw= 2wew
2
1
�2�2
1
e
wdw=4e2�2e�2 e2�e( )=2e2
x=�
2 dx=2� d�
�
� /2
�
�
3
cos �
2( ) d� =� �
� /2
�
2
cos �
2( )� 12 (2� d� )=
1
2 �
�
� /2
xcos x}{dx} u=x
dv=cos xdx du=dx v=sin x
1
2 �� /2
�
xcos xdx = 12 xsin x
�
� /2
� �
� /2
�
sin xdx = 12 xsin x+cos x
�
� /2
= 12 (� sin� +cos� )�
1
2
�
2 sin
�
2 +cos
�
2 =
1
2 (� � 0�1)�
1
2
�
2 � 1+0 =�
1
2 �
�
4
�x5ex
2
dx =� x
2( ) 2ex
2
xdx=�t2et 12 dt t=x
2
�
1
2 dt=xdx
=
1
2 t
2
�2t+2( ) et+C = 12 x
4
�2x
2
+2( ) ex
2
+C
So .
32. Let , , . Then
 . For , let , 
 , . So . Thus,
 .
33. Let , so that and . Thus, . Now use parts with
 , , , to get
34. Let , so that and . Thus, . Now use parts with ,
 , , to get .
35. Let , so that . Thus,
 . Now use parts with ,
 , , to get
36.
 [ where ]
 [ by Example 3] 
 7
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
u=x dv=cos� xdx�du=dx v= sin� x( ) /�
�xcos� xdx=x� sin� x� ��
sin� x
� dx=
xsin� x
� +
cos� x
�
2
+C
F f 0
u=ln x dv=x3/2dx�du= 1
x
dx v= 25 x
5/2
�x3/2ln xdx = 25 x
5/2
ln x�
2
5 �x
3/2dx= 25 x
5/2
ln x�
2
5
2
x
5/2
+C
= 25 x
5/2
ln x�
4
25 x
5/2
+C .
F f
u=2x+3 dv=exdx�du=2dx v=ex
�(2x+3)exdx=(2x+3)ex�2�exdx=(2x+3)ex�2ex+C= 2x+1( ) ex+C
F f
37. Let , , . Then
 . We see from the graph that this is
reasonable, since has extreme values where is .
38. Let , , . Then
We see from the graph that this is reasonable, since has a minimum where changes from negative
to positive.
39. Let , , . Then
 . We see from the graph that this is
reasonable, since has a minimum where changes from negative to positive.
 8
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�x3ex
2
dx=�x2� xex
2
dx=I
u=x
2 dv=xex
2
dx�du=2xdx v= 12 e
x
2
I=
1
2 x
2
e
x
2
��xex
2
dx= 12 x
2
e
x
2
�
1
2 e
x
2
+C=
1
2 e
x
2
x
2
�1( )+C
F f
n=2 �sin 2xdx=� 12 cos xsin x+
1
2 �1dx=
x
2 �
sin 2x
4 +C
�sin 4xdx=� 14 cos xsin
3
x+
3
4 �sin
2
xdx=� 14 cos xsin
3
x+
3
8 x�
3
16 sin 2x+C
u=cos
n�1
x dv=cos xdx�du=�(n�1)cos n�2xsin xdx v=sin x
�cos nxdx = cos n�1xsin x+(n�1)�cos n�2xsin 2xdx
= cos
n�1
xsin x+(n�1)�cos n�2x 1�cos 2x( )dx
= cos
n�1
xsin x+(n�1)�cos n�2xdx�(n�1)�cos nxdx
n�cos nxdx=cos n�1xsin x+(n�1)�cos n�2xdx
�cos nxdx= 1
n
cos
n�1
xsin x+
n�1
n
�cos n�2xdx
n=2 �cos 2xdx= 12 cos xsin x+
1
2 �1dx=
x
2 +
sin 2x
4 +C
�cos 4xdx= 14 cos
3
xsin x+
3
4 �cos
2
xdx= 14 cos
3
xsin x+
3
8 x+
3
16 sin 2x+C
�sin nxdx=� 1
n
cos xsin
n�1
x+
n�1
n
�sin n�2xdx
40. .
Let , , . Then
 . We see from the graph that this is
reasonable, since has a minimum where changes from negative to positive.
41. (a) Take in Example 6 to get .
(b) .
42. (a) Let , , in (2):
Rearranging terms gives or
(b) Take in part (a) to get .
(c)
43. (a) From Example 6, . Using (6),
 9
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�� /2
0
sin
n
xdx = � cos xsin
n�1
x
n
� /2
0
+
n�1
n
�� /2
0
sin
n�2
xdx
= (0�0)+
n�1
n
�� /2
0
sin
n�2
xdx= n�1
n
�� /2
0
sin
n�2
xdx
n=3 �� /2
0
sin
3
xdx= 23 �
� /2
0
sin xdx= � 23 cos x
� /2
0
=
2
3
n=5 �� /2
0
sin
5
xdx= 45 �
� /2
0
sin
3
xdx= 45 �
2
3 =
8
15
n=1 2n+1=3 k� 1
�� /2
0
sin
2k+1
xdx= 2� 4� 6� � � � � 2k( )3� 5� 7� � � � � 2k+1( )
�� /2
0
sin
2k+3
xdx = 2k+22k+3 �
� /2
0
sin
2k+1
xdx= 2k+22k+3 �
2� 4� 6� � � � � 2k( )
3� 5� 7� � � � � 2k+1( )
= 2� 4� 6� � � � � (2k)[2 k+1( ) ]3� 5� 7� � � � � (2k+1)[2 k+1( )+1] ,
n=k+1 n� 1
n=1
�� /2
0
sin
2
xdx= 12 �
� /2
0
1dx= 12 x
� /2
0
=
1
2 �
�
2
k� 1 �� /2
0
sin
2k
xdx= 1� 3� 5� � � � � (2k�1)2� 4� 6� � � � � 2k( )
�
2
�� /2
0
sin
2 k+1( )
xdx = 2k+12k+2 �
� /2
0
sin
2k
xdx= 2k+12k+2 �
1� 3� 5� � � � � (2k�1)
2� 4� 6� � � � � 2k( )
�
2
= 1� 3� 5� � � � � (2k�1)(2k+1)2� 4� 6� � � � � (2k)(2k+2) �
�
2 ,
n=k+1 n� 1
u=(ln x)
n dv=dx�du=n(ln x)n�1 dx/x( ) v=x
�(ln x)ndx=x(ln x)n��nx(ln x)n�1 dx/x( )=x(ln x)n�n�(ln x)n�1dx
u=x
n dv=exdx�du=nxn�1dx v=ex �xnexdx=xnex�n�xn�1exdx
u= x
2
+a
2( ) n dv=dx�du=n x2+a2( ) n�12xdx v=x
(b) Using in part (a), we have .
Using in part (a), we have .
(c) The formula holds for (that is, ) by (b). Assume it holds for some . Then
 . By Example 6,
so the formula holds for . By induction, the formula holds for all .
44. Using Exercise 43 (a), we see that the formula holds for , because
 .
Now assume it holds for some . Then . By Exercise 43(a),
so the formula holds for . By induction, the formula holds for all .
45. Let , , . By Equation 2,
 .
46. Let , , . By Equation 2, .
47. Let , , . Then
 10
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
� x2+a2( ) ndx =x x2+a2( ) n�2n�x2 x2+a2( ) n�1dx
x x
2
+a
2( ) n�2n � x2+a2( ) ndx�a2� x2+a2( ) n�1dx since x2= x2+a2( )�a2
� 2n+1( )� x2+a2( ) ndx=x x2+a2( ) n+2na2� x2+a2( ) n�1dx
� x2+a2( ) ndx= x x
2
+a
2( ) n
2n+1 +
2na
2
2n+1 � x
2
+a
2( ) n�1dx 2n+1�0
u=sec
n�2
x dv=sec 2xdx�du= n�2( )sec n�3xsec xtan xdx v=tan x
�sec nxdx = tan xsec n�2x� n�2( )�sec n�2xtan 2xdx
= tan xsec
n�2
x� n�2( )�sec n�2x sec 2x�1( )dx
= tan xsec
n�2
x� n�2( )�sec nxdx+ n�2( )�sec n�2xdx
n�1( )�sec nxdx=tan xsec n�2x+ n�2( )�sec n�2xdx n�1�0
�sec nxdx= tan xsec
n�2
x
n�1 +
n�2
n�1 �sec
n�2
xdx
n=3 �(ln x)3dx=x ln x( ) 3�3�(ln x)2dx=x(ln x)3�3x(ln x)2+6xln x�6x+C
n=2
n=4
�x4exdx = x4ex�4�x3exdx=x4ex�4 x3�3x2+6x�6( ) ex+C
= e
x
x
4
�4x
3
+12x
2
�24x+24( )+C
n=3 n=2 n=1
=�5
0
xe
�0.4xdx u=x dv=e�0.4xdx�
du=dx v=�2.5e�0.4x
area = �2.5xe
�0.4x 5
0
+2.5�5
0
e
�0.4xdx
=�12.5e
�2
+0+2.5 �2.5e
�0.4x 5
0
=�12.5e
�2
�6.25(e
�2
�1)=6.25�18.75e
�2 25
4 �
75
4 e
�2
=
 , and
 [ provided ].
48. Let , , . Then by Equation 2,
so . If , then
 .
49. Take in Exercise 45 to get [
by Exercise 13 ].
Or : Instead of using Exercise 13 , apply Exercise 45 again with .
50. Take in Exercise 46 to get
 [by Exercise 14 ]
Or: Instead of using Exercise 14 , apply Exercise 46 with , then , then .
51. Area . Let , 
 , . Then
 or 
 11
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
y=xln x y=5ln x xln x=5ln x�xln x�5ln x=0� x�5( ) ln x=0
x=1 x=5 1<x<5 5ln x>xln x ln x>0
=�5
1
5ln x�xln x( ) dx=�51 (5�x)ln x dx u=ln x dv= 5�x( ) dx�du=dx/x v=5x�
1
2 x
2
area = (ln x) 5x�
1
2 x
2 5
1
��5
1
5x�
1
2 x
2 1
x
dx=(ln 5) 252 �0��
5
1
5�
1
2 x dx
= 25
2 ln 5� 5x�
1
4 x
2 5
1
=
25
2 ln 5� 25�
25
4 � 5�
1
4 =
25
2 ln 5�14
y=xsin x y=(x�2)
2
a	1.04748 b	2.87307
=�b
a
xsin x�(x�2)
2 dx
= �xcos x+sin x�
1
3 (x�2)
3 b
a
	2.81358�0.63075=2.18283
y=arctan3x y=x/2 x=
a	
2.91379
area =�a
�a
arctan3x�
1
2 x dx=2�
a
0
arctan3x�
1
2 x dx
=2 xarctan3x�
1
6 ln (1+9x
2
)�
1
4 x
2 a
0
	2(1.39768)=2.79536
52. The curves and intersect when ; that is,
when or . For , we have since . Thus, area
 . Let , , . Then
53. The curves and intersect at and , so
area
 [ by Example 1]
54. The curves and intersect at , so
 [ see Example 5]
 .
 12
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
V =�1
0
2� xcos (� x/2)dx u=x dv=cos (� x/2)dx�du=dx v= 2� sin (� x/2)
V =2�
2
� xsin
� x
2
1
0
�2� �
2
� �
1
0
sin
� x
2 dx=2�
2
� �0 �4 �
2
� cos
� x
2
1
0
=4+
8
� (0�1)=4�
8
�
=�1
0
2� x e
x
�e
�x( )dx=2� �1
0
xe
x
�xe
�x( )dx
=2� �1
0
xe
xdx��1
0
xe
�xdx
=2� xe
x
�e
x( )� �xe�x�e�x( ) 1
0
=2� [2/e�0]=4� /e
=�0
�1
2� (1�x)e
�xdx u=1�x dv=e�xdx�du=�dx v=�e�x
V =2� (1�x)(�e
�x
)
0
�1
�2� �0
�1
e
�xdx=2� (x�1)(e�x)+e�x
0
�1
=2� xe
�x 0
�1
=2� (0+e)=2� e
Volume =��
1
2� y� ln ydy=2� 12 y
2
ln y�
1
4 y
2 �
1
=2�
1
4 y
2
(2ln y�1)
�
1
=2�
�
2
(2ln� �1)
4 �
(0�1)
4 =�
3
ln� �
�
3
2 +
�
2
f (x)=x2ln x 1,3 f
ave
=
1
3�1 �
3
1
x
2
ln xdx= 12 I
u=ln x dv=x2dx�du=(1/x)dx v= 13 x
3
55. . Let , , .
 .
56.
Volume
 [ both integrals by parts]
57. Volume . Let , , .
58.
59. The average value of on the interval is .
Let , , . So
 13
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
I=
1
3 x
3
ln x
3
1
��3
1
1
3 x
2dx= 9ln 3�0( ) � 19 x
3 3
1
=9ln 3� 3�
1
9 =9ln 3�
26
9
f
ave
=
1
2 I=
1
2 9ln 3�
26
9 =
9
2 ln 3�
13
9
H=�60
0
v(t)dt 60
H = �60
0
�gt�v
e
ln
m�rt
m
dt=�g 12 t
2 60
0
�v
e
�60
0
ln (m�rt)dt��60
0
ln mdt
= �g(1800)+v
e
(ln m)(60)�v
e
�60
0
ln (m�rt)dt
u=ln (m�rt) dv=dt�du= 1
m�rt (�r)dt v=t
�60
0
ln (m�rt)dt = tln (m�rt) 60
0
+�60
0
rt
m�rt dt=60ln (m�60r)+�
60
0
�1+
m
m�rt dt
= 60ln (m�60r)+ �t�
m
r
ln (m�rt)
60
0
= 60ln (m�60r)�60�
m
r
ln (m�60r)+
m
r
ln m
H=�1800g+60v
e
ln m�60v
e
ln (m�60r)+60v
e
+
m
r
v
e
ln (m�60r)�
m
r
v
e
ln m g=9.8
m=30 000 r=160 v
e
=3000 H	14 844
v(t)>0 t s(t)=�t
0
v(w)dw=�t
0
w
2
e
�wdw
u=w
2 dv=e�wdw�du=2wdw v=�e�w s(t)= �w2e�w
t
0
+2�t
0
we
�wdw
U =w dV =e�wdw�dU =dw V =�e�w
s(t) = �t
2
e
�t
+2 �we
�w t
0
+�t
0
e
�wdw =�t2e�t+2 �te�t+0+ �e�w
t
0
= �t
2
e
�t
+2 �te
�t
�e
�t
+1( )=�t2e�t�2te�t�2e�t+2
= 2�e
�t
t
2
+2t+2( )meters
f (0)=g(0)=0 u= f (x) dv=g / /(x)dx�du= f /(x)dx v=g /(x)
�a
0
f (x)g / /(x)dx= f (x)g /(x) a
0
��a
0
f /(x)g /(x)dx= f (a)g /(a)��a
0
f /(x)g /(x)dx U = f /(x)
 .
Thus, .
60. The rocket will have height after seconds.
Let , , . Then
So . Substituting ,
 , , , and gives us , m.
61. Since for all , the desired distance is .
First let , , . Then .
Next let , , . Then
62. Suppose and let , , . Then
 . Now let ,
 14
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
dV =g /(x)dx�dU = f / /(x)dx V =g(x)
�a
0
f /(x)g /(x)dx= f /(x)g(x) a
0
��a
0
f / /(x)g(x)dx= f /(a)g(a)��a
0
f / /(x)g(x)dx
�a
0
f (x)g / /(x)dx= f (a)g /(a)� f /(a)g(a)+�a
0
f / /(x)g(x)dx
I=�4
1
xf / /(x)dx u=x dv= f / /(x)dx�du=dx v= f /(x)
I= xf /(x) 4
1
��4
1
f /(x)dx=4 f /(4)�1� f /(1)�[ f (4)� f (1)]=4� 3�1� 5�(7�2)=12�5�5=2
f / / I
g(x)=x g
/
(x)=1
�b
a
f (x)dx=bf (b)�a f (a)��b
a
x f /(x)dx y= f (x) x=g(y)
dy= f /(x)dx �b
a
x f /(x)dx=� f b( )f a( ) g(y)dy
ABFC
= bf (b) � af (a) � � f b( )f a( ) g(y)dy
= area of rectangle OBFE( ) � area of rectangle OACD( ) � area of region DCFE( )
f (x)=ln x f �1(x)=ex g= f �1 g(y)=ey
�e
1
ln xdx=eln e�1ln 1��ln e
ln 1
e
ydy=e��1
0
e
ydy=e� ey
1
0
=e�(e�1)=1
=�d
0
� b2dy��c
0
� a
2dy��d
c
� g(y)
2dy=� b2d�� a2c��d
c
� g(y)
2dy y= f (x) dy= f /(x)dx
g(y)=x V =� b2d�� a2c�� �b
a
x
2 f /(x)dx u=x2
dv= f /(x)dx�du=2xdx v= f (x)
 and , so
 .
Combining the two results, we get .
63. For , let , , . Then
 .
We used the fact that is continuous to guarantee that exists.
64. (a) Take and in Equation 1.
(b) By part (a), . Now let , so that and
 . Then . The result follows.
(c) Part (b) says that the area of region is
(d) We have , so , and since , we have . By part (b),
 .
65. Using the formula for volumes of rotation and the figure, we see that Volume
 . Let , which gives 
and , so that . Now integrate by parts with , and
 , , and
 15
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�b
a
x
2 f /(x)dx= x2 f (x) b
a
��b
a
2x f (x)dx=b2 f (b)�a2 f (a)��b
a
2x f (x)dx f (a)=c f (b)=d�
V =� b2d�� a2c�� b2d�a2c��b
a
2xf (x)dx =�b
a
2� xf (x)dx
0� x�
�
2 0� sin x� 1 sin
2n+2
x� sin
2n+1
x� sin
2n
x
I
2n+2
� I
2n+1
� I
2n
I
2n+2
I
2n
=
1� 3� 5�� � � � 2(n+1)�1
2� 4� 6� � � � � 2(n+1)
�
2
1� 3� 5� � � � � (2n�1)
2� 4� 6� � � � � 2n( )
�
2
=
2(n+1)�1
2(n+1) =
2n+1
2n+2
I
2n
I
2n
I
2n+2
I
2n
�
I
2n+1
I
2n
�
I
2n
I
2n
2n+1
2n+2
2n+1
2n+2�
I
2n+1
I
2n
� 1 lim
n�
2n+1
2n+2 =limn�
1=1 lim
n�
I
2n+1
I
2n
=1
1 = lim
n�
I
2n+1
I
2n
=lim
n�
2� 4� 6� � � � � 2n( )
3� 5� 7� � � � � 2n+1( )
1� 3� 5� � � � � 2n�1( )
2� 4� 6� � � � � 2n( )
�
2
= lim
n�
2� 4� 6� � � � � 2n( )
3� 5� 7� � � � � 2n+1( )
2� 4� 6� � � � � 2n( )
1� 3� 5� � � � � 2n�1( )
2
�
= lim
n�
2
1 �
2
3 �
4
3 �
4
5 �
6
5 �
6
7 � � � � �
2n
2n�1 �
2n
2n+1 �
2
�
�
2
�
2 =
2
1 �
2
3 �
4
3 �
4
5 �
6
5 �
6
7 � � � �
k k 2n 2n�1 2n
 , but and 
 .
66. (a) We note that for , , so . So by the second
Comparison Property of the Integral, .
(b) Substituting directly into the result from Exercise 44 , we get
(c) We divide the result from part (a) by . The inequalities are preserved since is positive:
 . Now from part (b), the left term is equal to , so the expression becomes
 . Now , so by the Squeeze Theorem, .
(d) We substitute the results from Exercises 43 and 44 into the result from part (c):
 [rearrange terms]
Multiplying both sides by gives us the Wallis product :
(e) The area of the th rectangle is . At the th step, the area is increased from to by
 16
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
2n
2n�1 2n+1( ) 2n 2n+1
2n+1
2n
2n
2n�1
1
(2n+1)/(2n) =
2n
2n+1
2
1 �
2
3 �
4
3 �
4
5 �
6
5 �
6
7 � � � � =
�
2
multiplying the width by , and at the th step, the area is increased from to by
multiplying the height by . These two steps multiply the ratio of width to height by and
 respectively. So, by part (d), the limiting ratio is .
 17
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.1 Integration by Parts
�sin 3xcos 2xdx =�sin 2xcos 2xsin xdx=� 1�cos 2x( ) cos 2xsin xdx=� 1�u2( ) u2(�du)
=� u2�1( ) u2du=� u4�u2( ) du= 15 u
5
�
1
3 u
3
+C=
1
5 cos
5
x�
1
3 cos
3
x+C
�sin 6xcos 3xdx =�sin 6xcos 2xcos xdx=�sin 6x 1�sin 2x( )cos xdx=�u6 1�u2( )du
=� u6�u8( )du= 17 u
7
�
1
9 u
9
+C=
1
7 sin
7
x�
1
9 sin
9
x+C
�
� /2
3� /4
sin
5
xcos
3
xdx = �
� /2
3� /4
sin
5
xcos
2
xcos xdx= �
� /2
3� /4
sin
5
x 1�sin
2
x( ) cos xdx
s
= �
1
2 /2
u
5
1�u
2( ) du= �
1
2 /2
u
5
�u
7( ) du= 16 u
6
�
1
8 u
8 2 /2
1
=
1/8
6 �
1/16
8 �
1
6 �
1
8 =�
11
384
�
0
� /2
cos
5
xdx = �
0
� /2
cos
2
x( ) 2cos xdx= �
0
� /2
1�sin
2
x( ) 2cos xdx=�
0
1
1�u
2( ) 2du
=�
0
1
1�2u
2
+u
4( ) du= u� 23 u
3
+
1
5 u
5 1
0
= 1�
2
3 +
1
5 �0=
8
15
�cos 5xsin 4xdx =�cos 4xsin 4xcos xdx=� 1�sin 2x( ) 2sin 4xcos xdx=� 1�u2( ) 2u4du
=� 1�2u2+u4( )u4du=� u4�2u6+u8( )du= 15 u
5
�
2
7 u
7
+
1
9 u
9
+C
=
1
5 sin
5
x�
2
7 sin
7
x+
1
9 sin
9
x+C
1.
2.
3.
4.
5.
6.
 1
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
�sin 3(mx)dx =� 1�cos
2
mx( ) sin mxdx=� 1m � 1�u
2( ) du u=cos mx du=�msin mxdx
=�
1
m u�
1
3 u
3
+C=�
1
m cos mx�
1
3 cos
3
mx +C
=
1
3m cos
3
mx�
1
m cos mx+C
�� /2
0
cos
2
� d� =�� /2
0
1
2 (1+cos 2� )d� �
=
1
2 � +
1
2 sin 2�
� /2
0
=
1
2
�
2 +0 � 0+0( ) =
�
4
�� /2
0
sin
2
(2� )d� =�� /2
0
1
2 (1�cos 4� )d� =
1
2 � �
1
4 sin 4�
� /2
0
=
1
2
�
2 �0 � 0�0( ) =
�
4
��
0
sin
4
(3t)dt =��
0
sin
2
(3t)
2
dt=��
0
1
2 (1�cos 6t)
2
dt=
1
4 �
�
0
(1�2cos 6t+cos
2
6t)dt
=
1
4 �
�
0
1�2cos 6t+
1
2 (1+cos 12t) dt=
1
4 �
�
0
3
2 �2cos 6t+
1
2 cos 12t dt
=
1
4
3
2 t�
1
3 sin 6t+
1
24 sin 12t
�
0
=
1
4
3�
2 �0+0 � 0�0+0( ) =
3�
8
��
0
cos
6
� d� =��
0
(cos
2
� )
3
d� =��
0
1
2 (1+cos 2� )
3
d� =
1
8 �
�
0
(1+3cos 2� +3cos
2
2� +cos
3
2� )d�
=
1
8 � +
3
2 sin 2�
�
0
+
1
8 �
�
0
3
2 (1+cos 4� ) d� +
1
8 �
�
0
(1�sin
2
2� )cos 2� d�
=
1
8 � +
3
16 � +
1
4 sin 4�
�
0
+
1
8 �
0
0
(1�u
2
)
1
2 du u=sin 2� du=2cos 2� d�
=
�
8 +
3�
16 +0=
5�
16
 [ , ]
7.
 [ half angle identity]
8.
9.
10.
 [ , ]
11.
 2
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
�(1+cos� )2d� =�(1+2cos� +cos
2
� )d� =� +2sin� +
1
2 �(1+cos 2� )d�
=� +2sin� +
1
2 � +
1
4 sin 2� +C=
3
2 � +2sin� +
1
4 sin 2� +C
u=x dv=cos
2
xdx�du=dx v=�cos 2xdx=� 12 (1+cos 2x)dx=
1
2 x+
1
4 sin 2x
�xcos 2xdx =x
1
2 x+
1
4 sin 2x ��
1
2 x+
1
4 sin 2x dx=
1
2 x
2
+
1
4 xsin 2x�
1
4 x
2
+
1
8 cos 2x+C
=
1
4 x
2
+
1
4 xsin 2x+
1
8 cos 2x+C
�� /4
0
sin
4
xcos
2
xdx =�� /4
0
sin
2
x(sin xcos x)
2
dx=�� /4
0
1
2 (1�cos 2x)
1
2 sin 2x
2
dx
=
1
8 �
� /4
0
(1�cos 2x)sin
2
2xdx=
1
8 �
� /4
0
sin
2
2xdx�
1
8 �
� /4
0
sin
2
2xcos 2xdx
=
1
16 �
� /4
0
(1�cos 4x)dx�
1
16
1
3 sin
3
2x
� /4
0
=
1
16 x�
1
4 sin 4x�
1
3 sin
3
2x
� /4
0
=
1
16
�
4 �0�
1
3 =
1
192 (3� �4)
�� /2
0
sin
2
xcos
2
xdx =�� /2
0
1
4 4sin
2
xcos
2
x( )dx=�� /2
0
1
4 (2sin xcos x)
2
dx=
1
4 �
� /2
0
sin
2
2xdx
=
1
4 �
� /2
0
1
2 (1�cos 4x)dx=
1
8 �
� /2
0
(1�cos 4x)dx=
1
8 x�
1
4 sin 4x
� /2
0
=
1
8
�
2 =
�
16
�sin 3x cos x dx =� 1�cos 2x( ) cos xsin xdx=� 1�u2( )u1/2 �du( )=� u5/2�u1/2( )du
=
2
7 u
7/2
�
2
3 u
3/2
+C=
2
7 cos x( )
7/2
�
2
3 cos x( )
3/2
+C
=
2
7 cos
3
x�
2
3 cos x cos x +C
12. Let , , , so
13.
14.
15.
 3
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
u=sin� du=cos� d�
�cos� cos 5(sin� )d� =�cos
5
u du=�(cos 2u)2cos udu=�(1�sin 2u)2cos u du
=�(1�2sin 2u+sin 4u)cos u du=I
x=sin u dx=cos udu
I =�(1�2x
2
+x
4
)dx=x�
2
3 x
3
+
1
5 x
5
+C=sin u�
2
3 sin
3
u+
1
5 sin
5
u+C
=sin (sin� )�
2
3 sin
3
(sin� )+
1
5 sin
5
(sin� )+C
�cos 2xtan 3xdx =� sin
3
x
cos x dx=�
1�u
2( ) �du( )
u =�
�1
u +u du
=�ln u +
1
2 u
2
+C=
1
2 cos
2
x�ln cos x +C
�cot 5� sin 4� d� =� cos
5
�
sin
5
�
sin
4
� d� =� cos
5
�
sin� d� =�
cos
4
�
sin� cos� d� =�
1�sin
2
�( ) 2
sin� cos� d�
=� 1�u
2( ) 2
u du=�
1�2u
2
+u
4
u du=�
1
u �2u+u
3
du
=ln u �u
2
+
1
4 u
4
+C=ln sin� �sin
2
� +
1
4 sin
4
� +C
� 1�sin xcos x dx =� sec x�tan x( ) dx=ln sec x+tan x �ln sec x +C
by (1) and the boxed
formula above it
=ln sec x+tan x( ) cos x +C=ln 1+sin x +C
=ln 1+sin x( )+C 1+sin x� 0
� 1�sin xcos x dx =�
1�sin x
cos x �
1+sin x
1+sin x dx=�
1�sin
2
x( )dx
cos x 1+sin x( ) =�
cos xdx
1+sin x
16. Let . Then and
Now let . Then and
17.
18.
19.
 since 
Or:
 4
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
=� dww w=1+sin x dw=cos xdx
=ln w +C=ln 1+sin x +C=ln 1+sin x( )+C
�cos 2xsin 2xdx=2�cos 3xsin x dx=�2�u3du=� 12 u4
+C=�
1
2 cos
4
x+C
u=tan x du=sec
2
xdx �sec 2xtan xdx=�udu= 12 u
2
+C=
1
2 tan
2
x+C
v=sec x dv=sec xtan xdx �sec 2xtan xdx=�vdv= 12 v
2
+C=
1
2 sec
2
x+C
�� /2
0
sec
4
(t/2)dt =�� /4
0
sec
4
x(2dx) x=t/2 dx=
1
2 dt =2�
� /4
0
sec
2
x(1+tan
2
x)dx
=2�1
0
(1+u
2
)du u=tan x du=sec
2
xdx =2 u+
1
3 u
3 1
0
=2 1+
1
3 =
8
3
� tan 2xdx=� sec 2x�1( )dx=tan x�x+C
� tan 4xdx=� tan 2x sec 2x�1( )dx=� tan 2xsec 2xdx�� tan 2xdx= 13 tan
3
x�tan x+x+C
u=tan x
�sec 6t dt =�sec 4t� sec 2t dt=�(tan 2t+1)2sec 2t dt=�(u2+1)2du
=�(u4+2u2+1)du= 15 u
5
+
2
3 u
3
+u+C=
1
5 tan
5
t+
2
3 tan
3
t+tan t+C
�� /4
0
sec
4
� tan
4
� d� =�� /4
0
(tan
2
� +1)tan
4
� sec
2
� d� =�1
0
(u
2
+1)u
4
du
=�1
0
(u
6
+u
4
)du=
1
7 u
7
+
1
5 u
5 1
0
=
1
7 +
1
5 =
12
35
�� /3
0
tan
5
xsec
4
xdx =�� /3
0
tan
5
x(tan
2
x+1)sec
2
xdx
 [ where , ]
20.
21. Let , . Then .
Or: Let , . Then .
22.
 [ , ] 
 [ , ] 
23.
24.
(Set in the first integral and use Exercise 23 for the second.)
25.
26.
27.
 5
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
=� 3
0
u
5
(u
2
+1)du
=� 3
0
(u
7
+u
5
)du=
1
8 u
8
+
1
6 u
6 3
0
=
81
8 +
27
6 =
81
8 +
9
2 =
81
8 +
36
8 =
117
8
�� /3
0
tan
5
xsec
4
xdx =�� /3
0
tan
4
xsec
3
xsec xtan xdx=�� /3
0
(sec
2
x�1)
2
sec
3
xsec xtan xdx
=�2
1
(u
2
�1)
2
u
3
du
=�2
1
(u
4
�2u
2
+1)u
3
du=�2
1
(u
7
�2u
5
+u
3
)du
=
1
8 u
8
�
1
3 u
6
+
1
4 u
4 2
1
= 32�
64
3 +4 �
1
8 �
1
3 +
1
4 =
117
8
� tan 3(2x)sec 5(2x)dx =� tan 2(2x)sec 4(2x)� sec (2x)tan (2x)dx
=�(u2�1)u4( 12 du) u=sec (2x),du=2sec (2x)tan (2x)dx
=
1
2 �(u
6
�u
4
)du=
1
14 u
7
�
1
10 u
5
+C=
1
14 sec
7
(2x)�
1
10 sec
5
(2x)+C
� tan 3xsec xdx =� tan 2xsec xtan xdx=� sec 2x�1( )sec xtan xdx
=�(u2�1)du
=
1
3 u
3
�u+C=
1
3 sec
3
x�sec x+C
�� /3
0
tan
5
xsec
6
xdx =�� /3
0
tan
5
xsec
4
xsec
2
xdx=�� /3
0
tan
5
x 1+tan
2
x( ) 2sec 2xdx
=� 3
0
u
5
1+u
2( ) 2du u=tan x du=sec 2xdx =� 3
0
u
5
1+2u
2
+u
4( )du
=� 3
0
u
5
+2u
7
+u
9( )du= 16 u
6
+
1
4 u
8
+
1
10 u
10 3
0
=
27
6 +
81
4 +
243
10 =
981
20
Alternate solution:
28.
 [ ]
29.
30.
 [ , ] 
 6
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
�� /3
0
tan
5
xsec
6
xdx =�� /3
0
tan
4
xsec
5
xsec xtan xdx=�� /3
0
sec
2
x�1( ) 2sec 5xsec xtan xdx
=�2
1
u
2
�1( ) 2u5du u=sec x du=sec xtan xdx
=�2
1
u
4
�2u
2
+1( )u5du=�2
1
u
9
�2u
7
+u
5( )du
=
1
10 u
10
�
1
4 u
8
+
1
6 u
6 2
1
=
512
5 �64+
32
3 �
1
10 �
1
4 +
1
6 =
981
20
� tan 5xdx =� sec 2x�1( ) 2tan xdx=�sec 4xtan xdx�2�sec 2xtan xdx+� tan xdx
=�sec 3xsec xtan xdx�2� tan xsec 2xdx+� tan xdx
=
1
4 sec
4
x�tan
2
x+ln sec x +C
1
4 sec
4
x�sec
2
x+ln sec x +C
� tan 6aydy =� tan 4ay sec 2ay�1( )dy=� tan 4aysec 2aydy�� tan 4aydy
=
1
5a tan
5
ay�� tan 2ay sec 2ay�1( )dy
=
1
5a tan
5
ay�� tan 2aysec 2aydy+� sec 2ay�1( )dy
=
1
5a tan
5
ay�
1
3a tan
3
ay+
1
a tan ay�y+C
� tan
3
�
cos
4
�
d� =� tan 3� sec 4� d� =� tan 3� � (tan 2� +1)� sec 2� d�
=�u3(u2+1)duu=tan� du=sec 2� d�
=�(u5+u3)du= 16 u
6
+
1
4 u
4
+C=
1
6 tan
6
� +
1
4 tan
4
� +C
� tan 2xsec xdx =� sec 2x�1( )sec xdx=�sec 3xdx��sec xdx
Alternate solution:
 [ , ]
31.
 [ or ]
32.
33.
 , ]
34.
 7
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
=
1
2 sec xtan x+ln sec x+tan x( ) �ln sec x+tan x +C
=
1
2 sec xtan x�ln sec x+tan x( )+C
�� /2
� /6
cot
2
xdx=�� /2
� /6
csc
2
x�1( )dx= �cot x�x � /2
� /6
= 0�
�
2 � � 3�
�
6 = 3�
�
3
�� /2
� /4
cot
3
xdx =�� /2
� /4
cot x csc
2
x�1( )dx=�� /2
� /4
cot xcsc
2
xdx��� /2
� /4
cos x
sin x dx
= �
1
2 cot
2
x�ln sin x
� /2
� /4
= 0�ln 1( )� � 12 �ln
1
2
=
1
2 +ln
1
2
=
1
2 (1�ln 2)
�cot 3� csc 3� d� =�cot 2� csc 2� � csc� cot� d� =�(csc 2� �1)csc 2� � csc� cot� d�
=�(u2�1)u2� (�du) u=csc� du=�csc� cot� d�
=�(u2�u4)du= 13 u
3
�
1
5 u
5
+C=
1
3 csc
3
� �
1
5 csc
5
� +C
�csc 4xcot 6xdx =�cot 6x(cot 2x+1)csc 2xdx
=�u6(u2+1)� (�du) u=cot x du=�csc 2xdx
=�(�u8�u6)du=� 19 u
9
�
1
7 u
7
+C=�
1
9 cot
9
x�
1
7 cot
7
x+C
I=�csc xdx=� csc x csc x�cot x( )csc x�cot x dx=�
�csc xcot x+csc
2
x
csc x�cot x dx u=csc x�cot x�
du= �csc xcot x+csc
2
x( )dx I=�du/u=ln u =ln csc x�cot x +C
�csc 4xcot 6xdx =�cot 6x(cot 2x+1)csc 2xdx
=�u6(u2+1)� (�du) u=cot x du=�csc 2xdx
 [ by Example 8 and (1)]
35.
36.
37.
 [ , ]
38.
 [ , ]
39. . Let 
 . Then .
40.
 [ , ]
 8
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
=�(�u8�u6)du=� 19 u
9
�
1
7 u
7
+C=�
1
9 cot
9
x�
1
7 cot
7
x+C
�sin 5xsin 2xdx =� 12 cos (5x�2x)�cos (5x+2x) dx=
1
2 � cos 3x�cos 7x( ) dx
=
1
6 sin 3x�
1
14 sin 7x+C
�sin 3xcos xdx =� 12 sin (3x+x)+sin (3x�x) dx=
1
2 � sin 4x+sin 2x( ) dx
=�
1
8 cos 4x�
1
4 cos 2x+C
�cos 7� cos 5� d� =� 12 cos (7� �5� )+cos (7� +5� ) d� =
1
2 � cos 2� +cos 12�( ) d�
=
1
2
1
2 sin 2� +
1
12 sin 12� +C=
1
4 sin 2� +
1
24 sin 12� +C
� cos x+sin xsin 2x dx =
1
2 �
cos x+sin x
sin xcos x dx=
1
2 �(csc x+sec x)dx
=
1
2 ln csc x�cot x +ln sec x+tan x( )+C
� 1�tan
2
x
sec
2
x
dx=� cos 2x�sin 2x( )dx=�cos 2xdx= 12 sin 2x+C
� dxcos x�1
=� 1cos x�1 �
cos x+1
cos x+1 dx=�
cos x+1
cos
2
x�1
dx=� cos x+1
�sin
2
x
dx
=� �cot xcsc x�csc 2x( )dx=csc x+cot x+C
u=tan (t
2
)�du=2tsec
2
(t
2
)dt
41. Use Equation 2(b):
42. Use Equation 2(a):
43. Use Equation 2(c):
44.
 [ by Exercise 39 and (1)]
45.
46.
47. Let . Then
 9
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
�tsec 2 t2( ) tan 4 t2( ) dt=�u4 12 du =
1
10 u
5
+C=
1
10 tan
5
(t
2
)+C
u=tan
7
x dv=sec x tan xdx�du=7tan
6
xsec
2
xdx v=sec x
� tan 8xsec xdx =� tan 7x� sec x tan xdx=tan 7xsec x��7tan 6xsec 2xsec xdx
=tan
7
xsec x�7� tan 6x tan 2x+1( )sec xdx
=tan
7
xsec x�7� tan 8xsec xdx�7� tan 6xsec xdx
8� tan 8xsec xdx=tan 7xsec x�7� tan 6xsec xdx �� /4
0
tan
8
xsec x
dx=
1
8 tan
7
xsec x
� /4
0
�
7
8 �
� /4
0
tan
6
xsec xdx=
2
8 �
7
8 I
u=cos x�du=�sin xdx
�sin 5xdx = � 1�cos 2x( ) 2sin xdx=� 1�u2( ) 2 �du( )
= � �1+2u2�u4( )du=� 15 u
5
+
2
3 u
3
�u+C
= �
1
5 cos
5
x+
2
3 cos
3
x�cos x+C
F f x( ) >0
�sin 4xcos 4xdx =� 12 sin 2x
4
dx=
1
16 �sin
4
2xdx=
1
16 �
1
2 1�cos 4x( )
2
dx
=
1
64 � 1�2cos 4x+cos
2
4x( )dx
=
1
64 x�
1
2 sin 4x +
1
128 � 1+cos 8x( ) dx
 .
48. Let , , . Then
 .
Thus, and 
 .
49. Let . Then
Notice that is increasing when , so the graphs serve as a check on our work.
50.
 10
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
=
1
64 x�
1
2 sin 4x +
1128 x+
1
8 sin 8x +C
=
3
128 x�
1
128 sin 4x+
1
1024 sin 8x+C
f (x)=0 F
�sin 3xsin 6xdx =� 12 cos (3x�6x)�cos (3x+6x) dx
=
1
2 �(cos 3x�cos 9x)dx
=
1
6 sin 3x�
1
18 sin 9x+C
f (x)=0 F
�sec 4 x2 dx =� tan
2 x
2 +1 sec
2 x
2 dx
=� u2+1( )2du u=tan x2 du=
1
2 sec
2 x
2 dx
=
2
3 u
3
+2u+C=
2
3 tan
3 x
2 +2tan
x
2 +C
F f F
Notice that whenever has a horizontal tangent.
51.
Notice that whenever has a horizontal tangent.
52.
 [ , ]
Notice that is increasing and is positive on the intervals on which they are defined. Also, has
 11
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
f
f
ave =
1
2� �
�
��
sin
2
xcos
3
xdx=
1
2� �
�
��
sin
2
x 1�sin
2
x( )cos xdx
=
1
2� �
0
0
u
2
1�u
2( )du u=sin x
=0
u=cos x du=�sin xdx��sin xcos xdx=�u(�du)=� 12 u
2
+C=�
1
2 cos
2
x+C
1
u=sin x du=cos xdx��sin xcos xdx=�udu= 12 u
2
+C=
1
2 sin
2
x+C
2
�sin xcos xdx=� 12 sin 2xdx=�
1
4 cos 2x+C3
u=sin x dv=cos xdx du=cos xdx v=sin x
�sin xcos xdx=sin 2x��sin xcos xdx �sin xcos xdx= 12 sin
2
x+C
4
cos 2x=1�2sin
2
x=2cos
2
x�1 �
1
4 cos 2x=
1
2 sin
2
x�
1
4 =�
1
2 cos
2
x+
1
4
0<x<
�
2 0<sin x<1 sin
3
x<sin x
�� /2
0
sin x�sin
3
x( )dx=�� /2
0
sin x 1�sin
2
x( )dx=�� /2
0
cos
2
xsin xdx u=cos x�du=�sin xdx
=�0
1
u
2
�du( )=�10u
2
du=
1
3 u
3 1
0
=
1
3
sin x>0 0<x<
�
2 2sin
2
x�sin x 2sin x sin x�
1
2
sin x�
1
2 2sin
2
x�sin x
�
6 ,
�
2 0,
�
6
no horizontal tangent and is never zero.
53.
 [ where ]
54. (a) Let . Then .
(b) Let . Then .
(c)
(d) Let , . Then , ,
so , by Equation .1.2, so .
The answers differ from one another by constants. Since
 , we find that .
55. For , we have , so . Hence the area is
 . Now let .
Then area .
56. for , so the sign of [ which equals ] is the same
as that of . Thus is positive on and negative on . The
desired area is
 12
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
�� /6
0
sin x�2sin
2
x( )dx+�� /2
� /6
2sin
2
x�sin x( )dx
= �� /6
0
sin x�1+cos 2x( ) dx+�� /2� /6 1�cos 2x�sin x( ) dx
= �cos x�x+
1
2 sin 2x
� /6
0
+ x�
1
2 sin 2x+cos x
� /2
� /6
= �
3
2 �
�
6 +
3
4 � �1( )+
�
2 �
�
6 �
3
4 +
3
2
= 1+
�
6 �
3
2
�2�
0
cos
3
xdx=0 x �
sin x�
1
3 sin
3
x
2�
0
=0 sin x cos x
2n� n 0
�2
0
sin 2� xcos 5� xdx=0 x�
x�
�1
0
sin 2� xcos 5� xdx = 12 �
2
0
sin (2� x�5� x)+sin (2� x+5� x) dx
= 12 �
2
0
sin (�3� x)+sin 7� x dx
57.
It seems from the graph that , since the area below the axis and above the graph
looks about equal to the area above the axis and below the graph. By Example 1, the integral is
 . Note that due to symmetry, the integral of any odd power of or 
between limits which differ by ( any integer) is .
58.
It seems from the graph that , since each bulge above the axis seems to
have a corresponding depression below the axis. To evaluate the integral, we use a trigonometric
identity:
 13
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
= 1
2
1
3� cos (�3� x)�
1
7� cos 7� x
2
0
= 12
1
3� (1�1)�
1
7� (1�1) =0
V =��
� /2
� sin
2
xdx=� ��
� /2
1
2 (1�cos 2x)dx=�
1
2 x�
1
4 sin 2x
�
� /2
=�
�
2 �0�
�
4 +0 =
�
2
4
Volume =�� /4
0
� tan
2
x( ) 2dx=� �� /4
0
tan
2
x sec
2
x�1( )dx=� �� /4
0
tan
2
xsec
2
xdx�� �� /4
0
tan
2
xdx
=� �� /4
0
u
2
du�� �� /4
0
sec
2
x�1( )dx
=�
1
3 u
3 � /4
x=0
�� tan x�x
� /4
0
=�
1
3 tan
3
x�tan x+x
� /4
0
=�
1
3 �1+
�
4 =�
�
4 �
2
3
Volume =� �� /2
0
1+cos x( ) 2�12 dx=� �� /20 2cos x+cos
2
x( )dx
=� 2sin x+
1
2 x+
1
4 sin 2x
� /2
0
=� 2+
�
4 =2� +
�
2
4
Volume =� �� /2
0
1
2
� 1�cos x( ) 2 dx=� �� /20 2cos x�cos
2
x( )dx
=� 2sin x�
1
2 x�
1
4 sin 2x
� /2
0
=� 2�
�
4 �0 �0 =2� �
�
2
4
s= f (t)=�t
0
sin� ucos
2
� udu y=cos� u�dy=�� sin� udu
s=�
1
� �
cos� t
1
y
2
dy=�
1
�
1
3 y
3 cos� t
1
=
1
3� 1�cos
3
� t( )
E(t)
2
= 155sin (120� t)
2
=155
2
sin
2
(120� t)
E(t)
2
t=0 t=
1
60 60
59.
60.
61.
62.
63. . Let . Then
 .
64. (a) We want to calculate the square root of the average value of
 . First, we calculate the average value itself, by
integrating over one cycle (between and , since there are cycles per second) and
 14
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
1
60 �0
E(t)
2
ave
= 11/60 �
1/60
0
155
2
sin
2
(120� t) dt=60� 155
2�1/60
0
1
2 1�cos (240� t) dt
= 60� 155
2 1
2 t�
1
240� sin (240� t)
1/60
0
=60� 155
2 1
2
1
60 �0 � 0�0( ) =
155
2
2
155
2
�110
220= E(t)
2
ave
�
220
2 = E(t)
2
ave
=
1
1/60 �
1/60
0
A
2
sin
2
(120� t)dt=60A
2�1/60
0
1
2 [1�cos (240� t)]dt
= 30A
2
t�
1
240� sin (240� t)
1/60
0
=30A
2 1
60 �0 � 0�0( ) =
1
2 A
2
220
2
=
1
2 A
2
� A=220 2�311
m�n
��
��
sin mxcos nxdx = ��
��
1
2 sin (m�n)x+sin (m+n)x dx
= 1
2 �
cos (m�n)x
m�n �
cos (m+n)x
m+n
�
��
=0
m=n
��
��
sin mxsin nxdx=��
��
1
2 [cos (m�n)x�cos (m+n)x]dx m�n
1
2
sin (m�n)x
m�n �
sin (m+n)x
m+n
�
��
=0 m=n
��
��
1
2 [1�cos (m+n)x]dx=
1
2 x
�
��
�
sin (m+n)x
2(m+n)
�
��
=� �0=�
��
��
cos mxcos nxdx=��
��
1
2 cos (m�n)x+cos (m+n)x dx m�n
1
2
sin (m�n)x
m�n +
sin (m+n)x
m+n
�
��
=0 m=n
��
��
1
2 [1+cos (m+n)x]dx=
1
2 x
�
��
+
sin (m+n)x
2(m+n)
�
��
=� +0=�
dividing by :
The RMS value is just the square root of this quantity, which is V.
(b)
Thus, V.
65. Just note that the integrand is odd .
Or: If , calculate
If , then the first term in each set of brackets is zero.
66. . If , this is equal to
 . If , we get
 .
67. . If , this is equal to
 . If , we get
 .
 15
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
1
� �
�
��
f (x)sin mxdx=
1
� �
�
��
	m
n=1
ansin nx sin mx dx=	
m
n=1
an
� �
�
��
sin mxsin nxdx
m
am
� �� =am
68. . By
Exercise 66 , every term is zero except the th one, and that term is .
 16
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.2 Trigonometric Integrals
x=3sec� 0�� <
�
2 ��� <
3�
2 dx=3sec� tan� d�
x
2
�9 = 9sec
2
� �9 = 9 sec
2
� �1( ) = 9tan 2�
= 3 tan� =3tan� �
� 1
x
2
x
2
�9
dx=� 1
9sec
2
� � 3tan�
3sec� tan� d� =
1
9 �cos� d� =
1
9 sin� +C=
1
9
x
2
�9
x +C
�sec (� +� )=sec� ��� <
3�
2
x=3sin� �
�
2 �� �
�
2 dx=3cos� d�
9�x
2 = 9�9sin
2
� = 9 1�sin
2
�( ) = 9cos 2�
= 3 cos� =3cos�
�x3 9�x2 dx = �33sin 3� � 3cos� � 3cos� d� =35�sin 3� cos 2� d�
= 3
5�sin 2� cos 2� sin� d� =35� 1�cos 2�( )cos 2� sin� d�
= 3
5� 1�u2( ) u2(�du)[u=cos� ,du=�sin� d� ]
= 3
5� u4�u2( )du=35 15 u
5
�
1
3 u
3
+C=3
5 1
5 cos
5
� �
1
3 cos
3
� +C
= 3
5 1
5
9�x
2( ) 5/2
3
5
�
1
3
9�x
2( ) 3/2
3
3
+C
1. Let , where or . Then and
 for the relevant values of 
Note that , so the figure is sufficient forthe case .
2. Let , where . Then and
 for the relevant values of \italic{\theta}\,\,.
 1
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
= 15 9�x
2( ) 5/2�3 9�x2( ) 3/2+Cor� 15 x
2
+6( ) 9�x( ) 3/2+C
x=3tan� �
�
2 <� <
�
2 dx=3sec
2
� d�
x
2
+9 = 9tan
2
� +9 = 9 tan
2
� +1( ) = 9sec 2�
= 3 sec� =3sec� � .
� x
3
x
2
+9
dx = � 3
3
tan
3
�
3sec� 3sec
2
� d� =3
3� tan 3� sec� d� =33� tan 2� tan� sec� d�
= 3
3� sec 2� �1( ) tan� sec� d� =33� u2�1( )du[u=sec� ,du=sec� tan� d� ]
= 3
3 1
3 u
3
�u +C=3
3 1
3 sec
3
� �sec� +C=3
3 1
3
x
2
+9( ) 3/2
3
3
�
x
2
+9
3 +C
= 1
3 x
2
+9( ) 3/2�9 x2+9 +C or 13 x
2
�18( ) x2+9 +C
x=4sin� �� /2���� /2 dx=4cos� d�
16�x
2
= 16�16sin
2
� = 16cos
2
� =4 cos� =4cos� x=0 4sin� =0�� =0
x=2 3 4sin� =2 3� sin� =
3
2 �� =
�
3
3. Let , where . Then and
 for the relevant values of 
4. Let , where . Then and
 . When , , and when
 , . Thus, substitution gives
 2
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
�
0
2 3
x
3
16�x
2
dx
= �
0
� /3
4
3
sin
3
�
4cos� 4cos� d� =4
3 �
0
� /3
sin
3
� d�
=4
3 �
0
� /3
1�cos
2
�( )sin� d�
=�4
3 �
1
1/2
1�u
2( )du=�64 u� 13 u
3 1/2
1
=�64
1
2 �
1
24 � 1�
1
3 =�64 �
5
24 =
40
3
u=16�x
2
x
2
=16�u du=�2xdx
t=sec� dt=sec� tan� d� t= 2�� =
�
4 t=2�� =
�
3
�
2
2
1
t
3
t
2
�1
dt = �
� /4
� /3
1
sec
3
� tan�
sec� tan� d� =�
� /4
� /3
1
sec
2
�
d� = �
� /4
� /3
cos
2
� d�
= �
� /4
� /3 1
2 (1+cos 2� )d� =
1
2 � +
1
2 sin 2�
� /3
� /4
= 1
2
�
3 +
1
2
3
2 �
�
4 +
1
2 � 1 =
1
2
�
12 +
3
4 �
1
2 =
�
24 +
3
8 �
1
4
x=2tan� dx=2sec
2
� d� x=0�� =0 x=2�� =
�
4
�2
0
x
3
x
2
+4 dx =�� /40 2
3
tan
3
� � 2sec� � 2sec
2
� d� =2
5�� /4
0
tan
2
� sec
2
� sec� tan� d�
=2
5�� /4
0
(sec
2
� �1)sec
2
� sec� tan� d�
=2
5� 2
1
(u
2
�1)u
2
du [u=sec� ,du=sec� tan� d� ]
=2
5� 2
1
(u
4
�u
2
)du=2
5 1
5 u
5
�
1
3 u
3 2
1
=2
5 1
5 � 4 2 �
1
3 � 2 2 �
1
5 �
1
3
=32
2
15 2 +
2
15 =
64
15 2 +1( )
Or: Let , , .
5. Let , so , , and . Then
6. Let , so , , and . Then
 3
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
u=x
2
+4 x
2
=u�4 du=2xdx
x=5sin� dx=5cos� d�
� 1
x
2
25�x
2
dx = �
1
5
2
sin
2
� � 5cos�
5cos� d�
= 125 �csc
2
� d� =�
1
25 cot� +C
= �
1
25
25�x
2
x +C
x=asec� 0�� <
�
2 ��� <
3�
2 dx=asec� tan� d� x
2
�a
2
=atan�
� x
2
�a
2
x
4
dx = �
atan�
a
4
sec
4
�
asec� tan� d�
=
1
a
2
�sin 2� cos� d�
= 1
3a
2
sin
3
� +C= x
2
�a
2( ) 3/2
3a
2
x
3
+C
x=4tan� �
�
2 <� <
�
2 dx=4sec
2
� d�
Or: Let , , .
7. Let , so . Then
8. Let , where or . Then and ,
so
9. Let , where . Then and
 4
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
x
2
+16 = 16tan
2
� +16 = 16(tan
2
� +1)
= 16sec
2
� =4 sec�
=4sec� �
� dx
x
2
+16
=� 4sec
2
� d�
4sec� =�sec� d� =ln sec� +tan� +C1
=ln
x
2
+16
4 +
x
4 +C1=ln x
2
+16 +x �ln 4 +C
1
=ln x
2
+16 +x( )+C C=C
1
�ln 4
x
2
+16 +x>0
t= 2 tan� �
�
2 <� <
�
2 dt= 2sec
2
� d�
t
2
+2 = 2tan
2
� +2 = 2(tan
2
� +1) = 2sec
2
�
= 2 sec� = 2sec� �
� t
5
t
2
+2
dt =� 4 2 tan
5
�
2sec�
2sec
2
� d� =4 2� tan 5� sec� d� =4 2� sec 2� �1( ) 2sec� tan� d�
=4 2� u2�1( ) 2du u=sec� du=sec� tan� d� =4 2� u4�2u2+1( )du
 for the relevant values of .
 , where .
(Since , we don’t need the absolute value.)
10. Let , where . Then and
 for the relevant values of .
 [ , ] 
 5
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
=4 2
1
5 u
5
�
2
3 u
3
+u +C=
4 2
15 u(3u
4
�10u
2
+15)+C
=
4 2
15 �
t
2
+2
2
3� (t
2
+2)
2
2
2
�10
t
2
+2
2 +15 +C
=
4
15 t
2
+2 �
1
4 3(t
4
+4t
2
+4)�20(t
2
+2)+60 +C
=
1
15 t
2
+2 (3t
4
�8t
2
+32)+C
2x=sin� �
�
2 �� �
�
2 x=
1
2 sin� dx=
1
2 cos� d�
1�4x
2
= 1� 2x( ) 2 =cos�
� 1�4x2 dx = �cos�
1
2 cos� d� =
1
4 � 1+cos 2�( ) d�
= 14 � +
1
2 sin 2� +C=
1
4 (� +sin� cos� )+C
= 1
4 sin
�1
(2x)+2x 1�4x
2
+C
�1
0
x x
2
+4 dx=�5
4
u
1
2 du u=x
2
+4 du=2xdx =
1
2 �
2
3 u
3/2 5
4
=
1
3 5 5�8( )
x=3sec� 0�� <
�
2 ��� <
3�
2 dx=3sec� tan� d� x
2
�9 =3tan�
� x
2
�9
x
3
dx =� 3tan�
27sec
3
�
3sec� tan� d� =
1
3 �
tan
2
�
sec
2
�
d�
11. Let , where . Then , , and
 .
12. [ , ] 
13. Let , where or . Then and ,
so
 6
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
=
1
3 �sin
2
� d� =
1
3 �
1
2 (1�cos 2� )d� =
1
6 � �
1
12 sin 2� +C=
1
6 � �
1
6 sin� cos� +C
=
1
6 sec
�1 x
3 �
1
6
x
2
�9
x
3
x +C=
1
6 sec
�1 x
3 �
x
2
�9
2x
2
+C
u= 5 sin� du= 5cos� d�
� du
u 5�u
2 = �
1
5 sin� � 5 cos�
5 cos� d� =
1
5
�csc� d�
=
1
5
ln csc� �cot� +C [byExercise2.39
= 1
5
ln
5
u �
5�u
2
u +C
= 1
5
ln
5� 5�u
2
u +C
x=asin� �
�
2 ���
�
2 dx=acos� d�
x
2
dx
a
2
�x
2( ) 3/2 = �
a
2
sin
2
� acos� d�
a
3
cos
3
�
=� tan 2� d�
= � sec 2� �1( )d� =tan� �� +C
14. Let , so . Then
15. Let , where . Then and
 7
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
=
x
a
2
�x
2
�sin
�1 x
a +C
4x=3sec� 0�� <
�
2 ��� <
3�
2 dx=
3
4 sec� tan� d�
16x
2
�9 =3tan�
� dx
x
2
16x
2
�9
=�
3
4 sec� tan� d�
3
4
2
sec
2
� 3tan�
=
4
9 �cos� d� =
4
9 sin� +C=
4
9
16x
2
�9
4x +C=
16x
2
�9
9x +C
u=x
2
�7 du=2xdx � x
x
2
�7
dx=
1
2 �
1
u
du=
1
2 � 2 u +C= x
2
�7 +C
ax=bsec� ax( ) 2=b2sec 2� � ax( ) 2�b2=b2sec 2� �b2=b2 sec 2� �1( )=b2tan 2�
ax
2( )�b2 =btan� dx= ba sec� tan� d�
16. Let , where or . Then and
 , so
17. Let , so . Then .
18. Let , so . So
 , , and
 8
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
� dx
ax( ) 2�b2
3/2 = �
b
a sec� tan�
b
3
tan
3
�
d� = 1
ab
2
� sec�
tan
2
�
d� = 1
ab
2
� cos�
sin
2
�
d� = 1
ab
2
�csc� cot� d�
= �
1
ab
2
csc� +C=� 1
ab
2
ax
ax( ) 2�b2
+C=� x
b
2
ax( ) 2�b2
+C
x=tan� �
�
2 <� <
�
2 dx=sec
2
� d� 1+x
2
=sec�
� 1+x
2
x dx
=� sec�tan� sec
2
� d� =� sec�tan� (1+tan
2
� )d�
=�(csc� +sec� tan� )d�
=ln csc� �cot� +sec� +C
=ln
1+x
2
x �
1
x +
1+x
2
1 +C=ln
1+x
2
�1
x + 1+x
2
+C
t=5sin� �
�
2 �� �
�
2 dt=5cos� d� 25�t
2
=5cos�
� t
25�t
2
dt =� 5sin�5cos� 5cos� d� =5�sin� d�
=�5cos� +C=�5�
25�t
2
5 +C=� 25�t
2
+C
u=25�t
2
du=�2t dt
19. Let , where . Then and , so
 [ by Exercise 8.2.39]
20. Let , where . Then and , so
Or: Let , so .
 9
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitutionu=4�9x
2
�du=�18xdx x
2
=
1
9 4�u( )
�2/3
0
x
3
4�9x
2
dx = �04
1
9 (4�u)u
1/2
�
1
18 du=
1
162 �
4
0
4u
1/2
�u
3/2( )du
= 1
162
8
3 u
3/2
�
2
5 u
5/2 4
0
=
1
162
64
3 �
64
5 =
64
1215
3x=2sin� �
�
2 �� �
�
2
x=tan� �
�
2 <� <
�
2 dx=sec
2
� d� x
2
+1 =sec� x=0�� =0 x=1�
� =
�
4
�1
0
x
2
+1 dx = �� /40 sec� sec
2
� d� =�� /4
0
sec
3
� d�
= 12 sec� tan� +ln sec� +tan�
� /4
0
= 12 2� 1+ln 1+ 2( )�0�ln (1+0) =
1
2 2 +ln 1+ 2( )
5+4x�x
2
=�(x
2
�4x+4)+9=�(x�2)
2
+9 x�2=3sin� �
�
2 �� �
�
2 dx=3cos� d�
� 5+4x�x2 dx =� 9�(x�2)2 dx=� 9�9sin 2� 3cos� d�
=� 9cos 2� 3cos� d� =�9cos 2� d�
21. Let . Then and
Or: Let , where .
22. Let , where . Then , and , 
 , so
 [by Example 8.2.8]
23. . Let , , so . Then
 10
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
=
9
2 �(1+cos 2� )d� =
9
2 � +
1
2 sin 2� +C
=
9
2 � +
9
4 sin 2� +C=
9
2 � +
9
4 (2sin� cos� )+C
=
9
2 sin
�1 x�2
3 +
9
2 �
x�2
3 �
5+4x�x
2
3 +C
=
9
2 sin
�1 x�2
3 +
1
2 (x�2) 5+4x�x
2
+C
t
2
�6t+13= t
2
�6t+9( )+4= t�3( ) 2+22 t�3=2tan� dt=2sec 2� d�
� dt
t
2
�6t+13
= � 1
2tan�( ) 2+22
2sec
2
� d� =� 2sec
2
�
2sec� d�
= �sec� d� =ln sec� +tan� +C1
= ln
t
2
�6t+13
2 +
t�3
2 +C1
= ln t
2
�6t+13 +t�3 +CwhereC=C
1
�ln 2
9x
2
+6x�8= 3x+1( ) 2�9 u=3x+1 du=3dx � dx
9x
2
+6x�8
=�
1
3 du
u
2
�9
u=3sec�
24. . Let , so . Then
 [by Formula 8.2.1]
25. , so let , . Then . Now let
 , where
 11
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
0�� <
�
2 ��� <
3�
2 du=3sec� tan� d� u
2
�9 =3tan�
�
1
3 du
u
2
�9
= � sec� tan� d�3tan� =
1
3 �sec� d� =
1
3 ln sec� +tan� +C1=
1
3 ln
u+ u
2
�9
3 +C1
= 1
3 ln u+ u
2
�9 +C=
1
3 ln 3x+1+ 9x
2
+6x�8 +C
4x�x
2
=� x
2
�4x+4( )+4=4� x�2( ) 2 u=x�2 x=u+2 dx=du
� x
2
dx
4x�x
2
=� u+2( )
2
du
4�u
2
=� 2sin� +2( )
2
2cos� 2cos� d�
=4� sin 2� +2sin� +1( )d�
=2� 1�cos 2�( ) d� +8�sin� d� +4�d�
=2� �sin 2� �8cos� +4� +C
=6� �8cos� �2sin� cos� +C
=6sin
�1 1
2 u �4 4�u
2
�
1
2 u 4�u
2
+C
=6sin
�1 x�2
2 �4 4x�x
2
�
x�2
2 4x�x
2
+C
x
2
+2x+2=(x+1)
2
+1 u=x+1 du=dx
� dx
x
2
+2x+2( ) 2 = �
du
u
2
+1( ) 2
=� sec
2
� d�
sec
4
�
where u=tan� ,du=sec
2
� d� ,
and u
2
+1=sec
2
� x
= �cos 2� d� = 12 � 1+cos 2�( ) d� =
1
2 (� +sin� cos� )+C
 or . Then and , so
26. , so let . Then and , so
27. . Let , . Then
 12
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
=
1
2 tan
�1
u+ u
1+u
2
+C=
1
2 tan
�1
(x+1)+
x+1
x
2
+2x+2
+C
5�4x�x
2
=� x
2
+4x+4( )+9=9� x+2( ) 2 u=x+2�du=dx
� dx
5�4x�x
2( ) 5/2 =
� du
9�u
2( ) 5/2
=� 3cos� d�
(3cos� )
5
where u=3sin� , du=3cos� d� ,
and 9�u
2
=3cos�
= 181 �sec
4
� d� =
1
81 � tan
2
� +1( )sec 2� d� = 181
1
3 tan
3
� +tan� +C
=
1
243
u
3
9�u
2( ) 3/2
+ 3u
9�u
2
+C=
1
243
(x+2)
3
5�4x�x
2( ) 3/2
+ 3(x+2)
5�4x�x
2
+C
u=x
2
du=2xdx
�x 1�x4 dx =� 1�u
2 1
2 du =
1
2 �cos� � cos� d�
whereu=sin� ,du=cos� d� ,
and 1�u
2
=cos�
=
1
2 �
1
2 (1+cos 2� )d� =
1
4 � +
1
8 sin 2� +C=
1
4 � +
1
4 sin� cos� +C
=
1
4 sin
�1
u+
1
4 u 1�u
2
+C=
1
4 sin
�1
(x
2
)+
1
4 x
2
1�x
4
+C
u=sin t du=cos tdt
�
0
� /2
cos t
1+sin
2
t
dt =�
0
1
1
1+u
2
dt= �
0
� /4 1
sec� sec
2
� d�
where u=tan� , du=sec
2
� d� ,
and 1+u
2
=sec�
=�� /4
0
sec� d� = ln sec� +tan�
� /4
0
=ln 2 +1( )�ln (1+0)=ln 2 +1( )
x=atan� �
�
2 <� <
�
2 x
2
+a
2
=asec�
� dx
x
2
+a
2 = � asec
2
� d�
asec� =�sec� d� =ln sec� +tan� +C1=ln
x
2
+a
2
a +
x
a +C1
28. . Let . Then
29. Let , . Then
30. Let , . Then
 [ by (1) in Section 8.2]
31. (a) Let , where . Then and
 13
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
= ln x+ x
2
+a
2( )+C whereC=C
1
�ln a
x=asinh t dx=acosh t dt x
2
+a
2
=acosh t
� dx
x
2
+a
2
=� acosh t dtacosh t =t+C=sinh
�1 x
a +C
x=atan� �
�
2 <� <
�
2
I = � x
2
x
2
+a
2( ) 3/2
dx=� a
2
tan
2
�
a
3
sec
3
�
asec
2
� d� =� tan
2
�
sec� d� =�
sec
2
� �1
sec� d�
= � sec� �cos�( ) d� =ln sec� +tan� �sin� +C
= ln
x
2
+a
2
a +
x
a �
x
x
2
+a
2
+C=ln x+ x
2
+a
2( )� x
x
2
+a
2
+C
1
x=asinh t
I = � a
2
sinh
2
t
a
3
cosh
3
t
acosh t dt=� tanh 2t dt=� 1�sech2t( )dt=t�tanh t+C
= sinh
�1 x
a �
x
a
2
+x
2
+C
f (x)= x
2
�1 /x 1,7
1
7�1 �1
7
x
2
�1
x dx
=
1
6 �0
� tan�
sec� � sec� tan� d�
where x=sec� ,dx=sec� tan� d� ,
x
2
�1 =tan� ,and � =sec
�1
7
=
1
6 �
�
0
tan
2
� d� =
1
6 �
�
0
(sec
2
� �1)d�
=
1
6 tan� ��
�
0
=
1
6 (tan� �� )
=
1
6 48 �sec
�1
7( )
(b) Let , so that and . Then
 .
32. (a) Let , . Then
(b) Let . Then
33. The average value of on the interval is
 14
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
9x
2
�4y
2
=36� y=�
3
2 x
2
�4�
area = 2�3
2
3
2 x
2
�4 dx=3�3
2
x
2
�4 dx
= 3��
0
2tan� 2sec� tan� d�
where x=2sec� ,
dx=2sec� tan� d� ,
� =sec
�1 3
2
= 12��
0
sec
2
� �1( )sec� d� =12��
0
sec
3
� �sec�( )d�
= 12
1
2 sec� tan� +ln sec� +tan�( )�ln sec� +tan�
�
0
= 6 sec� tan� �ln sec� +tan�
�
0
= 6
3 5
4 �ln
3
2 +
5
2 =
9 5
2 �6ln
3+ 5
2
�POQ=
1
2 (rcos� )(rsin� )=
1
2 r
2
sin� cos� PQR=�rrcos� r
2
�x
2
dx
x=rcos u�dx=�rsin udu �� u�
�
2
� r2�x2 dx = �rsin u (�rsin u)du=�r
2�sin 2udu=� 12 r
2
(u�sin ucos u)+C
= �
1
2 r
2
cos
�1
(x/r)+
1
2 x r
2
�x
2
+C
= 1
2 �r
2
cos
�1
(x/r)+x r
2
�x
2 r
rcos�
= 12 0� �r
2
� +rcos� rsin�( )
34.
35. Area of . Area of region . Let
 for . Then we obtain
so
area of region
 15
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
= 12 r
2
� �
1
2 r
2
sin� cos�
area of sector POR( )= area of �POQ( )+ area of region PQR( )= 12 r
2
�
x= 2sec� 0�� <
�
2 ��� <
3�
2 dx= 2 sec� tan� d�
� dx
x
4
x
2
�2
=� 2sec� tan� d�
4sec
4
� 2 tan�
=
1
4 �cos
3
� d� =
1
4 � 1�sin
2
�( )cos� d�
=
1
4 sin� �
1
3 sin
3
� +C u=sin�
=
1
4
x
2
�2
x �
x
2
�2( ) 3/2
3x
3
+C
y=x
2
4�x
2
y=2�x x=0.81
x=2 x
2
4�x
2
>2�x 0.81,2( )
A	�2
0.81
x
2
4�x
2
�(2�x) dx=�2
0.81
x
2
4�x
2
dx� 2x�
1
2 x
2 2
0.81
x=2sin� �
�
2 �� �
�
2
and thus, .
36. Let , where or , so . Then
 [substitute ]
From the graph, it appears that our answer is reasonable.
37. From the graph, it appears that the curve and the line intersect at about 
and , with on . So the area bounded by the curve and the line is
 . To evaluate the integral, we put
 , where . Then
 16
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
dx=2cos� d� x=2�� =sin
�1
1=
�
2 x=0.81�� =sin
�1
0.405	0.417
�20.81
x
2
4�x
2
dx 	 �� /20.4174sin
2
� (2cos� )(2cos� d� )=4�� /2
0.417
sin
2
2� d� =4�� /2
0.417
1
2 (1�cos 4� )d�
= 2 � �
1
4 sin 4�
� /2
0.417
=2
�
2 �0 � 0.417�
1
4 (0.995) 	2.81
A	2.81� 2� 2�
1
2 � 2
2
� 2� 0.81�
1
2 � 0.81
2
	2.10
x=btan� dx=bsec
2
� d� x
2
+b
2
=bsec�
E P( ) =�
�a
L�a
�b
4��
0
x
2
+b
2( ) 3/2
dx= �b
4��
0
�
� 1
� 2 1
bsec�( ) 3
bsec
2
� d�
= �
4��
0
b
�
� 1
� 2 1
sec� d� =
�
4��
0
b
�
� 1
� 2
cos� d� =
�
4��
0
b
sin�
� 2
� 1
=
�
4��
0
b
x
x
2
+b
2
L�a
�a
=
�
4��
0
b
L�a
(L�a)
2
+b
2
+ a
a
2
+b
2
x
2
+y
2
=R
2
x
2
+(y�b)
2
=r
2
b= R
2
�r
2
 , , and . So
Thus, .
38. Let , so that and .
39. Let the equation of the large circle be . Then the equation of the small circle is
 , where is the distance between the centers of the circles. The desired area is
 17
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
A = �r
�r
b+ r
2
�x
2( )� R2�x2 dx=2�r
0
b+ r
2
�x
2
� R
2
�x
2( )dx
= 2�r
0
bdx+2�r
0
r
2
�x
2
dx�2�r
0
R
2
�x
2
dx
2br=2r R
2
�r
2
� a2�x2 dx = �a
2
cos
2
� d� [x=asin� ,dx=acos� d� ] =
1
2 a
2� 1+cos 2�( ) d�
= 12 a
2
� +
1
2 sin 2� +C=
1
2 a
2
(� +sin� cos� )+C
= a
2
2 arcsin
x
a +
a
2
2
x
a
a
2
�x
2
a +C=
a
2
2 arcsin
x
a +
x
2 a
2
�x
2
+C
A = 2r R
2
�r
2
+ r
2
arcsin(x/r)+x r
2
�x
2 r
0
� R
2
arcsin(x/R)+x R
2
�x
2 r
0
= 2r R
2
�r
2
+r
2 �
2 � R
2
arcsin(r/R)+r R
2
�r
2
=r R
2
�r
2
+
�
2 r
2
�R
2
arcsin(r/R)
�
�
A = 2�2
�5
25�y
2
dy
= 25 arcsin(y/5)+y 25�y
2 2
�5
y=5sin� ]
= 25 arcsin
2
5 +2 21 +
25
2 �	58.72ft
2
A
� (5)
2
	
58.72
25� 	0.748 74.8%
The first integral is just . To evaluate the other two integrals, note that
so the desired area is
40. Note that the circular cross sections of the tank are the same everywhere, so the percentage of the
total capacity that is being used is equal to the percentage of any cross section that is under water.
The underwater area is
so the fraction of the total capacity in use is or .
 18
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
R>r x
2
=r
2
�(y�R)
2
� x=� r
2
�(y�R)
2
g(y)=2 r
2
�(y�R)
2
V = �R+r
R�r
2� y� 2 r
2
�(y�R)
2
dy=�r
�r
4� (u+R) r
2
�u
2
du [where u=y�R]
= 4� �r
�r
u r
2
�u
2
du+4� R�r
�r
r
2
�u
2
du where u=rsin� ,du=rcos� d�
in the second tegral
= 4� �
1
3 r
2
�u
2( ) 3/2 r
�r
+4� R�� /2
�� /2
r
2
cos
2
� d� =�
4�
3 (0�0)+4� Rr
2�� /2
�� /2
cos
2
� d�
= 2� Rr
2�� /2
�� /2
(1+cos 2� )d� =2� Rr
2
� +
1
2 sin 2�
� /2
�� /2
=2�
2
Rr
2
V =8� R�r
0
r
2
�y
2
dy
y=rsin�
41. We use cylindrical shells and assume that . , so
 and
Another method: Use washers instead of shells, so as in Exercise 6.2.61(a) , but
evaluate the integral using .
 19
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.3 Trigonometric Substitution
2x
(x+3)(3x+1) =
A
x+3 +
B
3x+1
1
x
3
+2x
2
+x
=
1
x(x
2
+2x+1)
= 1
x(x+1)
2
=
A
x +
B
x+1 +
C
(x+1)
2
x�1
x
3
+x
2
=
x�1
x
2
(x+1)
=
A
x +
B
x
2
+
C
x+1
x�1
x
3
+x
=
x�1
x x
2
+1( ) =
A
x +
Bx+C
x
2
+1
2
x
2
+3x�4
=
2
(x+4)(x�1) =
A
x+4 +
B
x�1
x
2
+x+1
x
2
(x�1) x
2
+x+1( ) =
A
x�1 +
Bx+C
x
2
+x+1
x
3
x
2
+4x+3
=x�4+
13x+12
x
2
+4x+3
=x�4+
13x+12
(x+1)(x+3) =x�4+
A
x+1 +
B
x+3
2x+1
(x+1)
3
(x
2
+4)
2
=
A
x+1 +
B
(x+1)
2
+ C
(x+1)
3
+
Dx+E
x
2
+4
+ Fx+G
(x
2
+4)
2
x
4
x
4
�1
=
(x
4
�1)+1
x
4
�1
=1+
1
x
4
�1
=1+
1
(x
2
�1)(x
2
+1)
=1+
1
(x�1)(x+1)(x
2
+1)
=1+
A
x�1 +
B
x+1 +
Cx+D
x
2
+1
t
4
+t
2
+1
t
2
+1( ) t2+4( ) 2
=
At+B
t
2
+1
+
Ct+D
t
2
+4
+ Et+F
t
2
+4( ) 2
x
4
(x
3
+x)(x
2
�x+3)
=
x
4
x(x
2
+1)(x
2
�x+3)
=
x
3
(x
2
+1)(x
2
�x+3)
=
Ax+B
x
2
+1
+
Cx+D
x
2
�x+3
1. (a)
(b)
2. (a)
(b)
3. (a)
(b) is irreducible, so .
4. (a)
(b)
5. (a)
 [ or use long division] 
(b)
6. (a)
(b)
 1
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions
1
x
6
�x
3
=
1
x
3
x
3
�1( ) =
1
x
3
(x�1) x
2
+x+1( ) =
A
x +
B
x
2
+ C
x
3
+
D
x�1 +
Ex+F
x
2
+x+1
� xx�6 dx=�
(x�6)+6
x�6 dx=� 1+
6
x�6 dx=x+6ln x�6 +C
� r
2
r+4 dr =�
r
2
�16
r+4 +
16
r+4 dr=� r�4+
16
r+4 dr
=
1
2 r
2
�4r+16ln r+4 +C
x�9
(x+5)(x�2) =
A
x+5 +
B
x�2 (x+5)(x�2) x�9=A(x�2)+B(x+5)
2 x �7=7B�B=�1 �5 x �14=�7A�A=2
� x�9(x+5)(x�2) dx=�
2
x+5 +
�1
x�2 dx=2ln x+5 �ln x�2 +C
1
(t+4)(t�1) =
A
t+4 +
B
t�1 �1=A(t�1)+B(t+4)
t=1�1=5B� B=
1
5 t=�4�1=�5A� A=�
1
5
� 1(t+4)(t�1) dt=�
�1/5
t+4 +
1/5
t�1 dt=�
1
5 ln t+4 +
1
5 ln t�1 +C or
1
5 ln
t�1
t+4 +C
1
x
2
�1
=
1
(x+1)(x�1) =
A
x+1 +
B
x�1 (x+1)(x�1) 1=A(x�1)+B(x+1)
1 x 1=2B�B=
1
2 �1 x 1=�2A�A=�
1
2
�
2
3
1
x
2
�1
dx =�
2
3 �1/2
x+1 +
1/2
x�1 dx= �
1
2 ln x+1 +
1
2 ln x�1
3
2
= �
1
2 ln 4+
1
2 ln 2 � �
1
2 ln 3+
1
2 ln 1 =
1
2 (ln 2+ln 3�ln 4) or
1
2 ln
3
2
x�1
x
2
+3x+2
=
A
x+1 +
B
x+2 (x+1)(x+2) x�1=A(x+2)+B(x+1)
�2 x �3=�B� B=3 �1 x �2=A
7.
8.
 [ or use long division]
9. . Multiply both sides by to get .
Substituting for gives . Substituting for gives . Thus,
10. .
 . . Thus,
11. . Multiply both sides by to get .
Substituting for gives . Substituting for gives . Thus,
12. . Multiply both sides by to get .
Substituting for gives . Substituting for gives . Thus,
 2
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions
�
0
1
x�1
x
2
+3x+2
dx =�
0
1 �2
x+1 +
3
x+2 dx= �2ln x+1 +3ln x+2
1
0
=(�2ln 2+3ln 3)�(�2ln 1+3ln 2)=3ln 3�5ln 2 orln
27
32
� ax
x
2
�bx
dx=� axx(x�b) dx=�
a
x�b dx=aln x�b +C
a�b
1
(x+a)(x+b) =
1
b�a
1
x+a �
1
x+b a�b
� dx(x+a)(x+b) =
1
b�a ln x+a �ln x+b( )+C=
1
b�a ln
x+a
x+b +C
a=b � dx
(x+a)
2
=�
1
x+a +C
2x+3
(x+1)
2
=
A
x+1 +
B
x+1( ) 2
�2x+3=A(x+1)+B x=�1 B=1 x
A=2
�
0
1
2x+3
(x+1)
2
dx = �
0
1
2
x+1 +
1
x+1( ) 2
dx= 2ln (x+1)�
1
x+1
1
0
= 2ln 2�
1
2 � 2ln 1�1( )=2ln 2+
1
2
x
3
�4x�10
x
2
�x�6
=x+1+
3x�4
(x�3)(x+2)
3x�4
(x�3)(x+2) =
A
x�3 +
B
x+2 3x�4=A(x+2)+B(x�3)
x=3 x=�2 5=5A�A=1 �10=�5B�B=2
�
0
1
x
3
�4x�10
x
2
�x�6
dx =�
0
1
x+1+
1
x�3 +
2
x+2 dx=
1
2 x
2
+x+ln x�3 +2ln (x+2)
1
0
=
1
2 +1+ln 2+2ln 3 � 0+0+ln 3+2ln 2( )=
3
2 +ln 3�ln 2=
3
2 +ln
3
2
13.
14. If , , so if , then
If , then .
15. . Take to get , and equate coefficients of 
to get . Now
16. . Write . Then .
Taking and , we get and , so
 3
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration;7.4 Integration of Rational Functions by Partial Fractions
4y
2
�7y�12
y(y+2)(y�3) =
A
y +
B
y+2 +
C
y�3 � 4y
2
�7y�12=A(y+2)(y�3)+By(y�3)+Cy(y+2) y=0
�12=�6A A=2 y=�2 18=10B B=
9
5 y=3 3=15C C=
1
5
�
1
2
4y
2
�7y�12
y(y+2)(y�3) dy
= �
1
2 2
y +
9/5
y+2 +
1/5
y�3 dy= 2ln y +
9
5 ln y+2 +
1
5 ln y�3
2
1
= 2ln 2+
9
5 ln 4+
1
5 ln 1�2ln 1�
9
5 ln 3�
1
5 ln 2
= 2ln 2+
18
5 ln 2�
1
5 ln 2�
9
5 ln 3=
27
5 ln 2�
9
5 ln 3=
9
5 (3ln 2�ln 3)=
9
5 ln
8
3
x
2
+2x�1
x
3
�x
=
x
2
+2x�1
x(x+1)(x�1) =
A
x +
B
x+1 +
C
x�1 x(x+1)(x�1)
x
2
+2x�1=A(x+1)(x�1)+Bx(x�1)+Cx(x+1) 0 x �1=�A�A=1 �1
x �2=2B�B=�1 1 x 2=2C�C=1
� x
2
+2x�1
x
3
�x
dx=� 1x �
1
x+1 +
1
x�1 dx=ln x �ln x+1 +ln x�1 +C=ln
x(x�1)
x+1 +C
1
x+5( ) 2(x�1)
=
A
x+5 +
B
x+5( ) 2
+
C
x�1 �1=A(x+5)(x�1)+B(x�1)+C(x+5)
2
x=�5
1=�6B B=�
1
6 x=1 1=36C C=
1
36 x=�2
1=A(3)(�3)+B(�3)+C 3
2( )=�9A�3B+9C=�9A+ 12 +
1
4 =�9A+
3
4 9A=�
1
4 A=�
1
36
� 1
x+5( ) 2(x�1)
dx = �
�1/36
x+5 �
1/6
x+5( ) 2
+
1/36
x�1 dx
= �
1
36 ln x+5 +
1
6(x+5) +
1
36 ln x�1 +C
x
2
(x�3)(x+2)
2
=
A
x�3 +
B
x+2 +
C
(x+2)
2
� x
2
=A(x+2)
2
+B(x�3)(x+2)+C(x�3)
x=3 A=
9
25 x=�2 C=�
4
5 x
2
1=A+B
17. . Setting gives
 , so . Setting gives , so . Setting gives , so .
Now
18. . Multiply both sides by to get
 . Substituting for gives . Substituting 
for gives . Substituting for gives . Thus,
 .
19. . Setting gives
 , so . Setting gives , so . Setting gives
 , so and . Now
20. .
Setting gives . Take to get , and equate the coefficients of to get 
 4
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions
� B=
16
25
� x
2
(x�3)(x+2)
2
dx = �
9/25
x�3 +
16/25
x+2 �
4/5
(x+2)
2
dx
= 925 ln x�3 +
16
25 ln x+2 +
4
5(x+2) +C
5x
2
+3x�2
x
3
+2x
2
=
5x
2
+3x�2
x
2
x+2( )
=
A
x +
B
x
2
+
C
x+2 x
2
(x+2)
5x
2
+3x�2=Ax(x+2)+B(x+2)+Cx
2
x=�2 C=3 x=0 B=�1
x
2
5=A+C� A=2
� 5x
2
+3x�2
x
3
+2x
2
dx=� 2x �
1
x
2
+
3
x+2 dx=2ln x +
1
x +3ln x+2 +C
1
s
2
(s�1)
2
=
A
s +
B
s
2
+
C
s�1 +
D
(s�1)
2
�1=As(s�1)
2
+B(s�1)
2
+Cs
2
(s�1)+Ds
2
s=0 B=1
s=1 D=1 s
3
0=A+C A=�C s=2
1=2A+1�4A+4 A=2
� ds
s
2
(s�1)
2
=� 2s +
1
s
2
�
2
s�1 +
1
(s�1)
2
ds=2ln s �
1
s �2ln s�1 �
1
s�1 +C
x
2
(x+1)
3
=
A
x+1 +
B
(x+1)
2
+ C
(x+1)
3
x+1( ) 3 x2=A(x+1)2+B(x+1)+C
x=�1 C=1 x
2
A=1 x=0 B=�2
� x
2
dx
(x+1)
3
=� 1x+1 �
2
(x+1)
2
+ 1
(x+1)
3
dx=ln x+1 +
2
x+1 �
1
2(x+1)
2
+C
x
x+1 =
(x+1)�1
x+1 =1�
1
x+1
x
3
(x+1)
3
= 1�
1
x+1
3
=1�
3
x+1 +
3
(x+1)
2
� 1
(x+1)
3
� x
3
(x+1)
3
dx=� 1� 3x+1 +
3
(x+1)
2
� 1
(x+1)
3
dx=x�3ln x+1 �
3
x+1 +
1
2(x+1)
2
+C
 . Then
21. . Multiply by to get
 . Set to get , and take to get . Equating the
coefficients of gives . So
 .
22. . Set , giving .
Then set to get . Equate the coefficients of to get or , and finally set to
get or . Now
 .
23. . Multiply by to get . Setting
 gives . Equating the coefficients of gives , and setting gives . Now
 .
24. , so . Thus,
 .
 5
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions
10
(x�1) x
2
+9( ) =
A
x�1 +
Bx+C
x
2
+9
(x�1) x
2
+9( )
10=A x
2
+9( )+(Bx+C)(x�1) * 1 x 10=10A�A=1 0 x
10=9A�C�C=9(1)�10=�1 x
2
� * 0=A+B
� B=�1.
� 10
(x�1)(x
2
+9)
dx =� 1x�1 +
�x�1
x
2
+9
dx=� 1x�1 �
x
x
2
+9
�
1
x
2
+9
dx
=ln x�1 �
1
2 ln x
2
+9( ) [let u=x2+9] � 13 tan
�1 x
3 [Formula10] +C
x
2
�x+6
x
3
+3x
=
x
2
�x+6
x x
2
+3( ) =
A
x +
Bx+C
x
2
+3
x x
2
+3( ) x2�x+6=A x2+3( )+(Bx+C)x
0 x 6=3A�A=2 x
2
� 1=A+B�
B=1�2=�1 x � �1=C
� x
2
�x+6
x
3
+3x
dx =�
2
x +
�x�1
x
2
+3
dx=� 2x �
x
x
2
+3
�
1
x
2
+3
dx
=2ln x �
1
2 ln x
2
+3( )� 1
3
tan
�1 x
3
+C
x
3
+x
2
+2x+1
x
2
+1( ) x2+2( ) =
Ax+B
x
2
+1
+
Cx+D
x
2
+2
x
2
+1( ) x2+2( )
x
3
+x
2
+2x+1=(Ax+B) x
2
+2( )+(Cx+D) x2+1( )�
x
3
+x
2
+2x+1= Ax
3
+Bx
2
+2Ax+2B( )+ Cx3+Dx2+Cx+D( )�
x
3
+x
2
+2x+1=(A+C)x
3
+(B+D)x
2
+(2A+C)x+(2B+D)
A+C=1 B+D=1
2A+C=2 2B+D=1
A=1 C=0
B=0 D=1 I=� x
3
+x
2
+2x+1
x
2
+1( ) x2+2( ) dx=�
x
x
2
+1
+
1
x
2
+2
dx
25. . Multiply both sides by to get
 ( ). Substituting for gives . Substituting for gives
 . The coefficients of the terms in ( ) must be equal, so 
 Thus,
26. . Multiply by to get .
Substituting for gives . The coefficients of the terms must be equal, so 
 . The coefficients of the terms must be equal, so . Thus,
27. . Multiply both sides by to get
 . Comparing coefficients gives us the following
system of equations:
(1) (2)
(3) (4)
Subtracting equation (1) from equation (3) gives us , so . Subtracting equation (2) from
equation (4) gives us , so . Thus, . For
 6
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions
� x
x
2
+1
dx u=x
2
+1 du=2xdx � x
x
2
+1
dx=
1
2 �
1
u du=
1
2 ln u +C=
1
2 ln x
2
+1( )+C
� 1
x
2
+2
dx a= 2 � 1
x
2
+2
dx=� 1
x
2
+ 2( ) 2
dx=
1
2
tan
�1 x
2
+C
I=
1
2 ln x
2
+1( )+ 1
2
tan
�1 x
2
+C
x
2
�2x�1
x�1( ) 2 x2+1( ) =
A
x�1 +
B
(x�1)
2
+
Cx+D
x
2
+1
� x
2
�2x�1=A(x�1) x
2
+1( )+B x2+1( )+(Cx+D) x�1( ) 2
x=1 B=�1 x
3
A=�C
�1=�A�1+D D=A
x=2 �1=5A�5�2A+A A=1
� x
2
�2x�1
x�1( ) 2 x2+1( ) dx
= �
1
x�1 �
1
(x�1)
2
�
x�1
x
2
+1
dx
= ln x�1 +
1
x�1 �
1
2 ln x
2
+1( )+tan �1x+C
� x+4
x
2
+2x+5
dx =� x+1
x
2
+2x+5
dx+� 3
x
2
+2x+5
dx=
1
2 �
(2x+2)dx
x
2
+2x+5
+� 3dx
(x+1)
2
+4
=
1
2 ln x
2
+2x+5 +3� 2du
4(u
2
+1)
where x+1=2u,
and dx=2du
=
1
2 ln (x
2
+2x+5)+
3
2 tan
�1
u+C=
1
2 ln (x
2
+2x+5)+
3
2 tan
�1 x+1
2 +C
x
3
�2x
2
+x+1
x
4
+5x
2
+4
=
x
3
�2x
2
+x+1
x
2
+1( ) x2+4( ) =
Ax+B
x
2
+1
+
Cx+D
x
2
+4
� x
3
�2x
2
+x+1= Ax+B( ) x2+4( )+ Cx+D( ) x2+1( )
A+C=1 B+D=�2 4A+C=1 4B+D=1� A=0 C=1 B=1 D=�3
� x
3
�2x
2
+x+1
x
4
+5x
2
+4
dx=� dx
x
2
+1
+� x�3
x
2
+4
dx=tan
�1
x+
1
2 ln x
2
+4( )� 32 tan
�1
(x/2)+C
1
x
3
�1
=
1
(x�1) x
2
+x+1( ) =
A
x�1 +
Bx+C
x
2
+x+1
�1=A x
2
+x+1( )+(Bx+C)(x�1)
 , let so and then . For
 , use Formula 10 with . So . Thus,
 .
28. .
Setting gives . Equating the coefficients of gives . Equating the constant terms
gives , so ,
and setting gives or . We have
29.
30.
. Equating coefficients gives , , , , , , . Now
 .
31. .
 7
Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.4 Integration of Rational Functions by Partial Fractions
x=1 A=
1
3 x
2
0=
1
3 +B 1=
1
3 �C B=�
1
3 C=�
2
3 �
� 1
x
3
�1
dx = �
1
3
x�1 dx+�
�
1
3 x�
2
3
x
2
+x+1
dx=
1
3 ln x�1 �
1
3 �
x+2
x
2
+x+1