APS Questão 6

Prévia do material em texto

Lista de Exercícios - Cálculo Diferencial e Integral II 
 Prof. Dr. Marcelo Paraná e Monitor Ícaro Viterbre 
 
6) Resolva as integrais indefinidas pelos métodos de integração 
estudados: 
a) ∫ 𝑥 cos(2𝑥) 𝑑𝑥 
b) ∫ 𝑥 (−sen2(𝑥) + 1) 𝑑𝑥 
c) ∫
3𝑥−1
𝑥2−𝑥+1
 𝑑𝑥 
d) ∫
2𝑥−1
𝑥2−3𝑥+2 
 𝑑𝑥 
e) ∫ 3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 
f) ∫
𝑥−1
1+𝑥2
⋅ cos(2𝜋) 𝑑𝑥 
g) ∫ cos3(𝑥) 𝑑𝑥 
h) ∫
𝑒ln (𝑥)
1+𝑥4
 𝑑𝑥 
i) ∫ ln (𝑥 + 3) 𝑑𝑥 
j) ∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 
k) ∫ 𝑒ln(𝑥
2) ⋅ 2𝑥 𝑑𝑥 
 
Gabarito 
 
6) 
 
a) 
x ⋅sen(2x)
2
+ 
cos(2𝑥)
4
+ C 
b) 
x2
4
+ 
cos(2x)
8
+ 
x ⋅ sen (2x)
4
+ C 
c) 
3 ⋅ln|x2−x+1|
2
+ 
1
√3
 ⋅
arctg (
2x−1
√3
) + C 
d) ln |
(x−2)3
x−1
| + C 
e) 
−3x⋅ e−2x
2
− 
3⋅e−2x 
4
+ C 
f) 
ln|1+x2|
2
− arctg(x) + C 
g) 
 2 sen3(𝑥)+3 sen(𝑥) cos2(𝑥) 
3
+ C 
h) 
arctg (x2)
2
+ C 
i) (x + 3) ⋅ ln(x + 3) − x + C 
j) 
e2x
13
 ⋅ [3 ⋅ sen(3x) + 2 ⋅
cos(3x)] + C 
k) 
x4
2
+ C 
 
 
1 
 
 
a) 
∫𝑥 ∙ cos(2𝑥)𝑑𝑥 
 
𝑢 = 𝑥 𝑑𝑢 = 1 𝑑𝑥 
𝑑𝑣 = cos(2𝑥)𝑑𝑥 𝑣 = ∫𝑑𝑣 
 
∫cos(2𝑥)𝑑𝑥 
𝑤 = 2𝑥 𝑑𝑤 = 2 𝑑𝑥 𝑑𝑥 =
𝑑𝑤
2
 
∫cos(𝑤)
𝑑𝑤
2
 
sen(𝑤)
2
 ∴
sen(2𝑥)
2
 
𝑣 =
sen(2𝑥)
2
 
 
 
∫𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫𝑣 𝑑𝑢 
 
 
∫𝑥 ⋅ cos(2𝑥) 𝑑𝑥 = 𝑥 ⋅
sen(2𝑥)
2
− ∫
sen(2𝑥)
2
 𝑑𝑥 
 
∫𝑥 ⋅ cos(2𝑥) 𝑑𝑥 =
𝑥 ⋅ sen(2𝑥)
2
−
1
2
⋅ ∫ sen(2𝑥)𝑑𝑥 
 
∫sen(2𝑥) 𝑑𝑥 
ℎ = 2𝑥 𝑑ℎ = 2 𝑑𝑥 𝑑𝑥 =
𝑑ℎ
2
 
∫sen(ℎ)
𝑑ℎ
2
 
−
cos(ℎ)
2
 ∴ −
cos(2𝑥)
2
 
 
2 
 
 
∫𝑥 ⋅ cos(2𝑥) 𝑑𝑥 =
𝑥 ⋅ sen(2𝑥)
2
−
1
2
⋅ (−
cos(2𝑥)
2
) 
 
∫𝑥 ⋅ cos(2𝑥) 𝑑𝑥 =
𝑥 ⋅ sen(2𝑥)
2
+
cos(2𝑥)
4
+ 𝐶 
 
𝑥 ⋅ sen(2𝑥)
2
+
cos(2𝑥)
4
+ 𝐶 
 
 
b) 
∫𝑥 ⋅ (− sen2(𝑥) + 1) 𝑑𝑥 
 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑡𝑎𝑏𝑒𝑙𝑎 𝑑𝑒 𝑖𝑑𝑒𝑛𝑡𝑖𝑑𝑎𝑑𝑒𝑠 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚é𝑡𝑟𝑖𝑐𝑎𝑠, 𝑖𝑡𝑒𝑚 "1" . 
cos2(𝑥) + sen2(𝑥) = 1 ∴ cos2(𝑥) = − sen2(𝑥) + 1 
 
∫𝑥 ⋅ cos2(𝑥) 𝑑𝑥 
𝑢 = 𝑥 𝑑𝑢 = 1 𝑑𝑥 
𝑑𝑣 = cos2(𝑥) 𝑑𝑥 𝑣 = ∫𝑑𝑣 
 
∫cos2(𝑥) 𝑑𝑥 
 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑡𝑎𝑏𝑒𝑙𝑎 𝑑𝑒 𝑖𝑑𝑒𝑛𝑡𝑖𝑑𝑎𝑑𝑒𝑠 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚é𝑡𝑟𝑖𝑐𝑎𝑠, 𝑖𝑡𝑒𝑚 "5" . 
cos2(𝑥) =
1 + cos(2𝑥)
2
 
∫
1 + cos(2𝑥)
2
 𝑑𝑥 
∫
1
2
 𝑑𝑥 +∫
cos(2𝑥)
2
 𝑑𝑥 
∫
1
2
 𝑑𝑥 =
𝑥
2
 
∫
cos(2𝑥)
2
 𝑑𝑥 
 
3 
 
 
𝑤 = 2𝑥 𝑑𝑤 = 2 𝑑𝑥 𝑑𝑥 =
𝑑𝑤
2
 
∫
cos(𝑤)
2
𝑑𝑤
2
 
1
4
⋅ ∫ cos(𝑤)𝑑𝑤 
sen(𝑤)
4
 ∴ 
sen(2𝑥)
4
 
 
𝑣 =
𝑥
2
+
sen(2𝑥)
4
 
 
 
∫𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫𝑣 𝑑𝑢 
 
∫𝑥 ⋅ (− sen2(𝑥) + 1) 𝑑𝑥 = 𝑥 ⋅ (
𝑥
2
+
sen(2𝑥)
4
) − ∫(
sen(2𝑥)
4
+
𝑥
2
)𝑑𝑥 
∫𝑥 ⋅ (− sen2(𝑥) + 1) 𝑑𝑥 =
𝑥2
2
+
𝑥 ⋅ sen(2𝑥)
4
−
1
4
⋅ ∫ sen(2𝑥) 𝑑𝑥 − ∫
𝑥
2
 𝑑𝑥 
 
∫
𝑥
2
 𝑑𝑥 =
𝑥2
4
 
 
∫sen(2𝑥) 𝑑𝑥 
 
𝑡 = 2𝑥 𝑑𝑡 = 2 𝑑𝑥 𝑑𝑥 =
𝑑𝑡
2
 
∫sen(𝑡)
𝑑𝑡
2
 
1
2
⋅ ∫ sen(𝑡) 𝑑𝑡 
−
cos(𝑡)
2
 ∴ −
cos(2𝑥)
2
 
 
 
 
4 
 
 
∫𝑥 ⋅ (− sen2(𝑥) + 1) 𝑑𝑥 =
𝑥2
2
+
𝑥 ⋅ sen(2𝑥)
4
−
1
4
⋅ (−
cos(2𝑥)
2
) −
𝑥2
4
 
 
∫𝑥 ⋅ (− sen2(𝑥) + 1) 𝑑𝑥 =
𝑥2
2
−
𝑥2
4
+
𝑥 ⋅ sen(2𝑥)
4
+
cos(2𝑥)
8
 
 
𝑥2
2
−
𝑥2
4
=
𝑥2
4
 
 
∫𝑥 ⋅ (− sen2(𝑥) + 1) 𝑑𝑥 =
𝑥2
4
+
𝑥 ⋅ sen(2𝑥)
4
+
cos(2𝑥)
8
+ 𝐶 
 
𝑥2
4
+
𝑥 ⋅ sen(2𝑥)
4
+
cos(2𝑥)
8
+ 𝐶 
 
𝑥2
4
+
cos(2𝑥)
8
+
𝑥 ⋅ sen(2𝑥)
4
+ 𝐶 
 
 
c) 
∫
3𝑥 − 1
𝑥2 − 𝑥 + 1
 𝑑𝑥 
 
Δ = −3 
Δ ∉ ℝ 
 
𝑥2 − 𝑥 + 1 
𝑥2 − 2 (𝑥) − (
1
2
) + (
1
2
)
2
+
3
4
 
 
 
 
 
 
 
 
5 
 
 
(𝑥 −
1
2
)
2
+ (
√3
2
)
2
 
𝑢2 = (𝑥 −
1
2
)
2
 
𝑎2 = (
√3
2
)
2
 
 
𝑢 = 𝑥 −
1
2
 ∴ 𝑥 = 𝑢 +
1
2
 ∴ 𝑑𝑢 = 𝑑𝑥 
 
 
𝑥2 − 𝑥 + 1 = (𝑥 −
1
2
)
2
+ (
√3
2
)
2
 
 
∫
3𝑥 − 1
𝑥2 − 𝑥 + 1
 𝑑𝑥 = ∫
3𝑥 − 1
(𝑥 −
1
2)
2
+ (
√3
2 )
2 𝑑𝑥 
∫
3(𝑢 +
1
2) − 1
𝑢2 + 𝑎2
 𝑑𝑢 = ∫
3𝑢 +
3
2 − 1 
𝑢2 + 𝑎2
 𝑑𝑢 = ∫
3𝑢 +
1
2
𝑢2 + 𝑎2
𝑑𝑢 
 
∫
3𝑢 +
1
2
𝑢2 + 𝑎2
𝑑𝑢 = ∫
3𝑢
𝑢2 + 𝑎2
 𝑑𝑢 + ∫
1
2
𝑢2 + 𝑎2
 𝑑𝑢 
 
∫
3𝑢
𝑢2 + 𝑎2
 𝑑𝑢 = 3 ⋅ ∫
𝑢
𝑢2 + 𝑎2
 𝑑𝑢 
𝑤 = 𝑢2 + 𝑎2 𝑑𝑤 = 2𝑢 𝑑𝑢 𝑑𝑢 =
𝑑𝑤
2𝑢
 
3 ⋅ ∫
𝑢
𝑤
𝑑𝑤
2𝑢
 =
3
2
⋅ ∫
𝑑𝑤
𝑤
 
3
2
⋅ ln|𝑤| =
3
2
⋅ ln|𝑢2 + 𝑎2| =
3
2
⋅ ln |(𝑥 −
1
2
)
2
+ (
√3
2
)
2
| =
3 ⋅ ln|𝑥2 − 𝑥 + 1|
2
 
3 ⋅ ln|𝑥2 − 𝑥 + 1|
2
 
 
 
 
6 
 
 
∫
1
2
𝑢2 + 𝑎2
 𝑑𝑢 
 
∫
1
2
𝑢2 + 𝑎2
 𝑑𝑢 =
1
2
⋅ ∫
1
𝑢2 + 𝑎2
 𝑑𝑢 
 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑡𝑎𝑏𝑒𝑙𝑎 𝑑𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑖𝑠, 𝑖𝑡𝑒𝑚 "21" . 
∫
𝑑𝑥
𝑥2 + 𝑎2
=
arctg (
𝑥
𝑎)
𝑎
+ 𝐶 
 
1
2
⋅ ∫
1
𝑢2 + 𝑎2
 𝑑𝑢 =
1
2
⋅
1
𝑎
⋅ arctg (
𝑢
𝑎
) =
1
2
⋅
1
√3
2
⋅ arctg(
𝑥 −
1
2
√3
2
 ) 
 
1
2
⋅
1
√3
2
⋅ arctg(
𝑥 −
1
2
√3
2
 ) =
1
2
⋅
2
√3
⋅ arctg
(
 
 2
2
⋅ (
𝑥 −
1
2
√3
2
)
)
 
 
=
1
√3
⋅ arctg (
2𝑥 − 1 
√3
) 
 
∫
3𝑥 − 1
𝑥2 − 𝑥 + 1
 𝑑𝑥 =
3 ⋅ ln|𝑥2 − 𝑥 + 1|
2
+
1
√3
⋅ arctg (
2𝑥 − 1 
√3
) + 𝐶 
 
 
3 ⋅ ln|𝑥2 − 𝑥 + 1|
2
+
1
√3
⋅ arctg (
2𝑥 − 1 
√3
) + 𝐶 
 
 
d) 
∫
2𝑥 − 1
𝑥2 − 3𝑥 + 2
 𝑑𝑥 
 
𝑥2 − 3𝑥 + 2 
Δ = 𝑏2 − 4𝑎𝑐 
𝑎 = 1 𝑏 = −3 𝑐 = 2 
 
7 
 
 
Δ = (−3)2 − 4 ⋅ 1 ⋅ 2 
Δ = 1 
𝑥 =
−𝑏 ± √Δ
2𝑎
 
𝑥 =
−(−3) ± √1
2 ⋅ 1
 
𝑥 =
3 ± 1
2
 
𝑥′ = 1 𝑥′′ = 2 
 
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 𝑎(𝑥 − 𝑥′)(𝑥 − 𝑥′′) 
 
2𝑥 − 1
𝑥2 − 3𝑥 + 2
= 
2𝑥 − 1
1 ⋅ (𝑥 − 2)(𝑥 − 1)
=
2𝑥 − 1
(𝑥 − 2)(𝑥 − 1)
 
2𝑥 − 1
(𝑥 − 2)(𝑥 − 1)
=
𝐴
𝑥 − 2
+
𝐵
𝑥 − 1
 
 
𝑃𝑎𝑟𝑎 𝑥 = 2 
(2𝑥 − 1)(𝑥 − 2)
(𝑥 − 2)(𝑥 − 1)
=
𝐴(𝑥 − 2)
𝑥 − 2
+
𝐵(𝑥 − 2)
𝑥 − 1
 
2𝑥 − 1
𝑥 − 1
= 𝐴 +
𝐵(𝑥 − 2)
𝑥 − 1
 
2 ⋅ 2 − 1
2 − 1
= 𝐴 +
𝐵(2 − 2)
2 − 1
 
 
𝐴 = 3 
 
𝑃𝑎𝑟𝑎 𝑥 = 1 
(2𝑥 − 1)(𝑥 − 1)
(𝑥 − 2)(𝑥 − 1)
=
𝐴(𝑥 − 1)
𝑥 − 2
+
𝐵(𝑥 − 1)
𝑥 − 1
 
2𝑥 − 1
𝑥 − 2
= 𝐵 +
𝐴(𝑥 − 1)
𝑥 − 2
 
2 ⋅ 1 − 1
1 − 2
= 𝐵 +
𝐴(1 − 1)
1 − 2
 
 
𝐵 = −1 
 
8 
 
 
2𝑥 − 1
(𝑥 − 2)(𝑥 − 1)
=
𝐴
𝑥 − 2
+
𝐵
𝑥 − 1
=
3
𝑥 − 2
+
−1
𝑥 − 1
 
 
∫
2𝑥 − 1
𝑥2 − 3𝑥 + 2
 𝑑𝑥 =∫(
3
𝑥 − 2
+
−1
𝑥 − 1
)𝑑𝑥 = ∫
3
𝑥 − 2
 𝑑𝑥 − ∫
𝑥
𝑥 − 1
 𝑑𝑥 
 
∫
3
𝑥 − 2
 𝑑𝑥 
𝑢 = 𝑥 − 2 𝑑𝑢 = 𝑑𝑥 
∫
3
𝑢
 𝑑𝑢 = 3 ⋅ ∫
1
𝑢
 𝑑𝑢 = 3 ⋅ ln|𝑢| 
3 ⋅ ln|𝑢| = 3 ⋅ ln|𝑥 − 2| 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑡𝑎𝑏𝑒𝑙𝑎 𝑑𝑒 𝑝𝑟𝑜𝑝𝑟𝑖𝑒𝑑𝑎𝑑𝑒𝑠 𝑑𝑜𝑠 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑜𝑠, 𝑖𝑡𝑒𝑚 "3" . 
3 ⋅ ln|𝑥 − 2| = ln|(𝑥 − 2)3| 
ln|(𝑥 − 2)3| 
 
 
∫
𝑥
𝑥 − 1
 𝑑𝑥 
𝑤 = 𝑥 − 1 𝑑𝑤 = 𝑑𝑥 
∫
1
𝑤
 𝑑𝑤 = ln|𝑤| 
ln|𝑤| = ln|𝑥 − 1| 
 
 
∫
2𝑥 − 1
𝑥2 − 3𝑥 + 2
 𝑑𝑥 = ln|(𝑥 − 2)3| − ln|𝑥 − 1| + 𝐶 
 
ln|(𝑥 − 2)3| − ln|𝑥 − 1| 
𝑂𝑏𝑠𝑒𝑟𝑣𝑎𝑟 𝑡𝑎𝑏𝑒𝑙𝑎 𝑑𝑒 𝑝𝑟𝑜𝑝𝑟𝑖𝑒𝑑𝑎𝑑𝑒𝑠 𝑑𝑜𝑠 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑜𝑠, 𝑖𝑡𝑒𝑚 "2" . 
ln|(𝑥 − 2)3| − ln|𝑥 − 1| = ln |
(𝑥 − 2)3
𝑥 − 1
| 
∫
2𝑥 − 1
𝑥2− 3𝑥 + 2
 𝑑𝑥 = ln |
(𝑥 − 2)3
𝑥 − 1
| + 𝐶 
 
 
9 
 
 
 
ln |
(𝑥 − 2)3
𝑥 − 1
| + 𝐶 
 
 
e) 
∫3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 
 
𝑢 = 3𝑥 𝑑𝑢 = 3 𝑑𝑥 
𝑑𝑣 = 𝑒−2𝑥 𝑣 = ∫𝑑𝑣 
 
∫𝑒−2𝑥 𝑑𝑥 
𝑤 = −2𝑥 𝑑𝑤 = −2 𝑑𝑥 𝑑𝑥 = −
𝑑𝑤
2
 
∫𝑒𝑤 ⋅ (−
𝑑𝑤
2
) = −
1
2
∫𝑒𝑤 𝑑𝑤 = −
𝑒𝑤
2
= −
𝑒−2𝑥
2
 
 
𝑣 = −
𝑒−2𝑥
2
 
 
 
 
∫𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫𝑣 𝑑𝑢 
 
∫3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 = −
3𝑥 ⋅ 𝑒−2𝑥
2
− ∫−
3 ⋅ 𝑒−2𝑥
2
 𝑑𝑥 
∫3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 = −
3𝑥 ⋅ 𝑒−2𝑥
2
+
3
2
⋅ ∫𝑒−2𝑥 𝑑𝑥 
 
 
 
 
 
 
10 
 
 
∫𝑒−2𝑥 𝑑𝑥 
𝑡 = −2𝑥 𝑑𝑡 = −2 𝑑𝑥 𝑑𝑥 = −
𝑑𝑢
2
 
∫𝑒𝑡 ⋅ (−
𝑑𝑡
2
) = −
1
2
⋅ ∫𝑒𝑡 𝑑𝑡 = −
𝑒𝑡
2
 
−
𝑒𝑡
2
= −
𝑒−2𝑥
2
 
 
−
𝑒−2𝑥
2
 
 
∫3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 = −
3𝑥 ⋅ 𝑒−2𝑥
2
+
3
2
⋅ (−
𝑒−2𝑥
2
) 
∫3𝑥 ⋅ 𝑒−2𝑥 𝑑𝑥 = −
3𝑥 ⋅ 𝑒−2𝑥
2
−
3 ⋅ 𝑒−2𝑥
4
+ 𝐶 
 
−
3𝑥 ⋅ 𝑒−2𝑥
2
−
3 ⋅ 𝑒−2𝑥
4
+ 𝐶 
 
 
f) 
∫
𝑥 − 1
1 + 𝑥2
⋅ cos(2𝜋) 𝑑𝑥 
 
cos(2𝜋) = 1 
 
∫
𝑥 − 1
1 + 𝑥2
⋅ 1 𝑑𝑥 = ∫
𝑥 − 1
1 + 𝑥2
 𝑑𝑥 
 
∫
𝑥 − 1
1 + 𝑥2
 𝑑𝑥 = ∫
𝑥
1 + 𝑥2
 𝑑𝑥 − ∫
1
1 + 𝑥2
 𝑑𝑥 
 
∫
1
1 + 𝑥2
 𝑑𝑥 = arctg(𝑥) 
 
 
11 
 
 
∫
𝑥
1 + 𝑥2
 𝑑𝑥 
𝑢 = 1 + 𝑥2 𝑑𝑢 = 2𝑥 𝑑𝑥 𝑑𝑥 =
𝑑𝑢
2𝑥
 
∫
𝑥
𝑢
𝑑𝑢
2𝑥
= ∫
𝑑𝑢
2𝑢
=
1
2
⋅ ∫
𝑑𝑢
𝑢
=
ln|𝑢|
2
 
 
ln|𝑢|
2
=
ln|1 + 𝑥2|
2
 
 
∫
𝑥 − 1
1 + 𝑥2
⋅ cos(2𝜋) 𝑑𝑥 =
ln|1 + 𝑥2|
2
− arctg(𝑥) + 𝐶 
 
ln|1 + 𝑥2|
2
− arctg(𝑥) + 𝐶 
 
 
g) 
∫cos3(𝑥) 𝑑𝑥 
 
∫cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 
 
𝑢 = cos2(𝑥) 𝑑𝑢 = −2 cos(𝑥) ⋅ sen(𝑥) 
 
𝑑𝑣 = cos(𝑥)𝑑𝑥 𝑣 = sen(𝑥) 
 
 
∫𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫𝑣 𝑑𝑢 
 
∫cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) − ∫−2 cos(𝑥) ⋅ sen(𝑥) ⋅ sen(𝑥) 𝑑𝑥 
 
∫cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) + ∫2 cos(𝑥) ⋅ sen2(𝑥) 𝑑𝑥 
 
∫cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) + 2 ⋅ ∫ cos(𝑥) ⋅ sen2(𝑥) 𝑑𝑥 
 
 
 
12 
 
 
∫cos(𝑥) ⋅ sen2(𝑥) 𝑑𝑥 
 
𝑡 = sen(𝑥) 𝑑𝑡 = cos(𝑥) 𝑑𝑥 
 
𝑑𝑥 =
𝑑𝑡
cos(𝑥)
 
 
∫cos(𝑥) ⋅ 𝑡2 ⋅
𝑑𝑡
cos(𝑥)
 
 
∫𝑡2 𝑑𝑡 
 
𝑡3
3
=
sen3(𝑥)
3
 
 
sen3(𝑥)
3
 
 
∫cos2(𝑥) ⋅ cos(𝑥) 𝑑𝑥 = cos2(𝑥) ⋅ sen(𝑥) + 2 ⋅
sen3(𝑥)
3
 
 
 
cos2(𝑥) ⋅ sen(𝑥) + 2 ⋅
sen3(𝑥)
3
+ 𝐶 
 
cos2(𝑥) ⋅ sen(𝑥) +
2 sen3(𝑥)
3
+ 𝐶 
 
3 cos2(𝑥) ⋅ sen(𝑥) + 2 sen3(𝑥)
3
+ 𝐶 
 
2 sen3(𝑥) + 3 sen (𝑥) ⋅ cos2(𝑥)
3
+ 𝐶 
 
 
h) 
∫
𝑒ln(𝑥)
1 + 𝑥4
 𝑑𝑥 𝑒ln(𝑎) = 𝑎 ∫
𝑥
1 + 𝑥4
 𝑑𝑥 
 
∫
𝑥
2 + (𝑥2)2
 𝑑𝑥 
 
𝑢 = 𝑥2 𝑑𝑢 = 2𝑥 𝑑𝑥 
 
𝑑𝑥 =
𝑑𝑢
2𝑥
 
 
 
13 
 
 
∫
𝑥
1 + 𝑢2
𝑑𝑢
2𝑥
 
 
∫
1
1 + 𝑢2
𝑑𝑢
2
 
 
1
2
∫
𝑑𝑢
1 + 𝑢2
 
 
1
2
⋅ arctg(𝑢) + 𝐶 
 
1
2
⋅ arctg(𝑥2) + 𝐶 
 
arctg(𝑥2)
2
+ 𝐶 
 
 
i) 
∫ln(𝑥 + 3) 𝑑𝑥 
 
𝑢 = 𝑥 + 3 𝑑𝑢 = 𝑑𝑥 
 
∫ln(𝑢) 𝑑𝑢 
 
∫ln(𝑢) 𝑑𝑢 = 𝑢 ⋅ ln(𝑢) − 𝑢 + 𝐶 
 
𝑢 ⋅ ln(𝑢) − 𝑢 + 𝐶 = (𝑥 + 3) ⋅ ln(𝑥 + 3) − (𝑥 + 3) + 𝐶 
 
(𝑥 + 3) ⋅ ln(𝑥 + 3) − (𝑥 + 3) + 𝐶 = (𝑥 + 3) ⋅ ln(𝑥 + 3) − 𝑥 − 3 + 𝐶 
 
−3 + 𝐶 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 ∴ −3 + 𝐶 = 𝐶 
 
 
(𝑥 + 3) ⋅ ln(𝑥 + 3) − 𝑥 + 𝐶 
 
 
j) 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 
 
𝑢 = 𝑒2𝑥 𝑑𝑢 = 2𝑒2𝑥 𝑑𝑥 
 
𝑑𝑣 = cos(3𝑥)𝑑𝑥 𝑣 = ∫𝑑𝑣 
 
14 
 
 
∫cos(3𝑥)𝑑𝑥 
 
𝑤 = 3𝑥 𝑑𝑤 = 3 𝑑𝑥 𝑑𝑥 =
𝑑𝑤
3
 
 
∫cos(𝑤) 
𝑑𝑤
3
 
 
sen(𝑤)
3
 
 
sen(3𝑥)
3
 
 
 
∫𝑢 𝑑𝑣 = 𝑢 ⋅ 𝑣 − ∫𝑣 𝑑𝑢 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥 ⋅ sen(3𝑥)
3
−∫
sen(3𝑥) ⋅ 2𝑒2𝑥
3
 𝑑𝑥 
 
−∫
sen(3𝑥) ⋅ 2𝑒2𝑥
3
 𝑑𝑥 
 
−
2
3
⋅ ∫ sen (3𝑥) ⋅ 𝑒2𝑥 𝑑𝑥 
 
𝑢 = 𝑒2𝑥 𝑑𝑢 = 2𝑒2𝑥 
 
𝑑𝑣 = sen(3𝑥) 𝑑𝑥 𝑣 = ∫𝑑𝑣 
 
 
∫sen(3𝑥) 𝑑𝑥 
 
ℎ = 3𝑥 𝑑ℎ = 3 𝑑𝑥 𝑑𝑥 =
𝑑ℎ
3
 
 
∫sen(ℎ)
𝑑ℎ
3
 
 
−
cos(ℎ)
3
 
 
−
cos(3𝑥)
3
 
 
 
15 
 
 
 
−
2
3
⋅ ∫ sen(3𝑥) ⋅ 𝑒2𝑥𝑑𝑥 = −
2
3
⋅ (−
𝑒2𝑥 ⋅ cos(3𝑥)
3
) −
2
3
⋅ (−∫−
2𝑒2𝑥 ⋅ cos(3𝑥)
3
 𝑑𝑥) 
 
2𝑒2𝑥 ⋅ cos(3𝑥)
9
−∫
4𝑒2𝑥 ⋅ cos(3𝑥)
9
 𝑑𝑥 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥 ⋅ sen(3𝑥)
3
−∫
sen (3𝑥) ⋅ 2𝑒2𝑥
3
 𝑑𝑥 
 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
− ∫
4𝑒2𝑥 ⋅ cos(3𝑥)
9
 𝑑𝑥 
 
 
 
 
∫
4𝑒2𝑥 ⋅ cos(3𝑥)
9
 𝑑𝑥 + ∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
4
9
⋅ ∫ 𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 +∫𝑒2𝑥 ⋅ cos(3𝑥)𝑑𝑥 =
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
4
9
+ 1 =
13
9
 
 
13
9
⋅ ∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
9
13
⋅ [
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
2𝑒2𝑥 ⋅ cos(3𝑥)
9
] 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
9
13
⋅ 
𝑒2𝑥 ⋅ sen(3𝑥)
3
+
9
13
⋅
2𝑒2𝑥 ⋅ cos(3𝑥)
9
 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 = 
3 𝑒2𝑥 ⋅ sen(3𝑥)
13
+
2𝑒2𝑥 ⋅ cos(3𝑥)
13
 
 
∫𝑒2𝑥 ⋅ cos(3𝑥) 𝑑𝑥 =
𝑒2𝑥
13
⋅ [3 ⋅ sen(3𝑥) + 2 ⋅ cos(3𝑥)] + 𝐶 
 
𝑒2𝑥
13
⋅ [3 ⋅ sen(3𝑥) + 2 ⋅ cos(3𝑥)] + 𝐶 
 
 
 
 
 
16 
 
 
 
k) 
∫𝑒ln(𝑥
2) ⋅ 2𝑥 𝑑𝑥 
 
𝑒ln(𝑎) = 𝑎 
 
∫𝑥2 ⋅ 2𝑥 𝑑𝑥 
 
∫2𝑥3𝑑𝑥 =
2𝑥4
4
+ 𝐶 
 
2𝑥4
4
+ 𝐶 =
𝑥4
2
+ 𝐶 
 
𝑥4
2
+ 𝐶