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Chapter 1 Techniques of Integration In this chapter we learn the most common integration techniques. These tech- niques will be explained throughout examples and remarks. 1.1 Integration by substitution We solve the following problems: 1. ∫ x(x2 + 4)6dx Solution: Since the derivative of x2+4 is 2x which may be considered as x, we do the substitution x2+4 = t⇒ 2xdx = dt⇒ dx = dt 2x . Therefore,∫ x(x2 + 4)6dx = ∫ xt6 dt 2x = 1 2 ∫ t6dt = 1 14 t7 + C = 1 14 (x2 + 4)7 + C. Thus, the idea of integration by substitution is that some quantity and its derivative both exist. In such a case we let the quantity (not its derivative) equal t. 2. ∫ x+1 x2+2x+7 dx 1 Solution: Let x2+2x+7 = t⇒ 2(x+1)dx = dt⇒ dx = dt 2(x+1) . Hence,∫ x+ 1 x2 + 2x+ 7 dx = ∫ x+ 1 t dt 2(x+ 1) = 1 2 ∫ 1 t dt = 1 2 ln |t|+ C = 1 2 ln |x2 + 2x+ 7|+ C. 3. ∫ sin √ x√ x dx Solution: Let √ x = t⇒ x = t2 ⇒ dx = 2tdt. Whence,∫ sin √ x x dx = ∫ sin t t 2tdt = 2 ∫ sin t d t = −2 cos t+ C = −2 cos√x+ C. 4. ∫ 1 (2x−1)5dx Solution 2x− 1 = t⇒ 2dx = dt⇒ dx = dt 2 . Therefore,∫ 1 (2x− 1)5dx = ∫ 1 t5 dt 2 = 1 2 ∫ t−5dt = 1 2 t−4 −4 + C = −1 8 1 (2x− 1)4 + C. 5. ∫ esin θ cos θdθ Solution: sin θ = t⇒ cos θdθ = dt⇒ dθ = dt cos θ . Whence,∫ esin θ cos θdθ = ∫ et cos θ dt cos θ = ∫ etdt == et + C = esin θ + C. 6. ∫ √ 5− tdt Solution: 5− t = u⇒ −dt = du⇒ dt = −du. Thus,∫ √ u(−du) = − ∫ u1/2du = −2 3 u3/2 + C = −2 3 (5− t)3/2 + C. 2 7. ∫ x2(x3 + 5)9dx Solution: x3 + 5 = t⇒ 3x2dx = dt⇒ dx = dt 3x2 . Thus,∫ x2(x3 + 5)9dx = ∫ x2t9 dt 3x2 = 1 3 ∫ t9dt = 1 3 t10 10 + C = 1 30 (x3 + 5)10 + C. 8. ∫ sin pitdt Solution: pit = x⇒ dt = dx pi ,∫ sin pitdt = ∫ sin x dx pi = 1 pi (− cosx) + C = − 1 pi cos pit+ C. 9. ∫ (lnx)2 x dx Solution: Let ln x = t⇒ 1 x dx = dt⇒ dx = xdt. Thus∫ (lnx)2 x dx = ∫ t2 x xdt = ∫ t2dt = t3 3 + C = 1 3 (lnx)3 + C. 10. ∫ cos θ sin6 θdθ Solution: sin θ = t⇒ cos θdθ = dt⇒ dθ = dt cos θ . Consequently∫ cos θ sin6 θdθ = ∫ cos θt6 dt cos θ = ∫ t6dt = t7 7 + C = sin7 θ 7 + C. 11. ∫ ex √ 1 + exdx Solution: 1 + ex = t⇒ exdx = dt⇒ dx = dt ex . Hence,∫ ex √ 1 + exdx = ∫ ex √ t dt ex = ∫ √ tdt = 2 3 t3/2 + c = 2 3 (1 + ex)3/2 + C. 3 12. ∫ 1 x lnx dx Solution: lnx = t⇒ 1 x dx = dt⇒ dx = xdt. Thus,∫ 1 x lnx dx = ∫ 1 xt xdt = ∫ 1 t dt = ln |t|+ C = ln | lnx|+ C. 13. ∫ √ cotx csc2 xdx Solution: cotx = t⇒ − csc2 xdx = dt⇒ dx = dt− csc2 x . Therefore∫ √ cotx csc2 xdx = ∫ √ t csc2 x dt − csc2 x = − ∫ √ tdt = −2 3 t3/2 + C = −2 3 (cotx)3/2 + C. 14. ∫ cotxdx Solution: We observe that∫ cotxdx = ∫ cosx sinx dx = ∫ 1 t dt = ln |t|+ C = ln | sin x|+ C. 15. ∫ sec3 x tan xdx Solution: sec x = t⇒ sec x tan xdx = dt⇒ dx = dt secx tanx . Therefore∫ sec3 x tan xdx = ∫ t3 tan x dt t tan x = ∫ t2dt = t3 3 + C = sec3 x 3 + C. 16. ∫ xa √ b+ cxa+1dx Solution: We see that the derivative of b+cxa+1 is (a+1)cxa. Thus the substitution b+ cxa+1 = t⇒ (a+ 1)cxadx = dt⇒ dx = dt (a+1)cxa leads to∫ xa √ b+ cxa+1dx = ∫ xa √ t dt (a+ 1)cxa = 1 c(a+ 1) ∫ √ tdt = 1 c(a+ 1) 2 3 t3/2 + C = 2 3c(a+ 1) (b+ cxa+1)3/2 + C. 4 17. ∫ sin t sec2(cos t)dt Solution: cos t = u⇒ − sin tdt = du⇒ dt = − du sin t . Thus,∫ sin t sec2(cos t)dt = ∫ sin t sec2(u) −du sin t = − ∫ sec2 udu = − tanu+ C = − tan(cos t) + C. 18. ∫ 1+x 1+x2 dx Solution: An immediate substitution will not simplify the problem. A deep look suggests that we split the top as follows:∫ 1 + x 1 + x2 dx = ∫ 1 1 + x2 dx+ ∫ x 1 + x2 dx = arctan(x) + 1 2 ln(x2 + 1) + C, where we have used the fact that d dx (arctanx) = 1 x2+1 for the first integral and the substitution x2+1 = t for the second one. The details are simple and shall be left to the student. 19. ∫ x 4√x+2dx Solution: Let 4 √ x+ 2 = t⇒ x+ 2 = t4 ⇒ dx = 4t3dt. Thus,∫ x 4 √ x+ 2 dx = ∫ t4 − 2 t 4t3dt = 4 ∫ t2(t4 − 2)dt = 4 ∫ (t6 − 2t2)dt = 4(t 7 7 − 2t 3 3 ) + C = 4 ( (x+ 2)7/4 7 − 2(x+ 2) 3/4 3 ) + C. 20. ∫ arctanx 1+x2 dx Solution: Let arctanx = t⇒ 1 x2+1 dx = dt⇒ dx = (x2 + 1)dt. Hence,∫ arctanx 1 + x2 dx = ∫ t 1 + x2 (x2 + 1)dt = ∫ tdt = t2 2 + C = (arctanx)2 2 + C. 21. ∫ √ x sin(1 + x3/2)dx 5 Solution: 1 + x3/2 = t⇒ 3 2 x1/2dx = dt⇒ dx = 2 3 √ x dt. Hence,∫ √ x sin(1 + x3/2)dx = ∫ √ x sin(t) 2 3 √ x dt = 2 3 ∫ sin tdt = −2 3 cos t+ C = −2 3 cos(1 + x3/2) + C. 22. ∫ (1 + tan θ)5 sec2 θdθ Solution: 1 + tan θ = t⇒ sec2 θdθ = dt⇒ dθ = dt sec2 θ . Thus∫ (1 + tan θ)5 sec2 θdθ = ∫ t5 sec2 θ dt sec2 θ = ∫ t5dt = t6 6 + C = (1 + tan θ)6 6 + C. 23. ∫ ex ex+1 dx Solution: ex + 1 = t⇒ exdx = dt⇒ dx = dt ex . Therefore,∫ ex ex + 1 dx = ∫ ex t dt ex = ∫ 1 t dt = ln |t|+ C = ln(ex + 1) + C. 24. ∫ cos(pi/x) x2 dx Solution: Let pi/x = t⇒ − pi x2 dx = dt⇒ dx = −x2dt pi . Therefore,∫ cos(pi/x) x2 dx = − ∫ cos t x2 x2dt pi = − 1 pi ∫ cos tdt = − 1 pi sin t+ C = − 1 pi sin(pi/x) + C. 25. ∫ sinx 1+cos2 x dx Solution: Let cos x = t⇒ − sin xdx = dt⇒ dx = − dt sinx . Whence,∫ sin x 1 + cos2 x dx = − ∫ sin x 1 + t2 dt sin x = ∫ −1 1 + t2 dt = − arctan t+ C = − arctan(cos x) + C. 26. ∫ x5 3 √ x3 + 1dx 6 Solution: Let x3 + 1 = t⇒ 3x2dx = dt⇒ dx = dt 3x2 . Thus,∫ x5 3 √ x3 + 1dx = ∫ x5 3 √ t dt 3x2 = 1 3 ∫ x3 3 √ tdt = 1 3 ∫ (t− 1)t1/3dt = 1 3 ∫ (t4/3 − t1/3)dt = 1 3 ( t7/3 7/3 − t 4/3 4/3 ) + C = 1 3 ( 3 7 (x3 + 1)7/3 − 3 4 (x3 + 1)4/3 ) + C. 27. ∫ x x4+1 dx Solution: Let x2 = t⇒ 2xdx = dt⇒ dx = dt 2x . Thus,∫ x x4 + 1 = ∫ x t2 + 1 dt 2x = 1 2 ∫ 1 t2 + 1 dt = 1 2 arctan t+ C = 1 2 arctan(x2) + C. 28. ∫ x2√ 1−xdx Solution: Let √ 1− x = t⇒ 1− x = t2 ⇒ dx = −2tdt. Consequently,∫ x2√ 1− xdx = ∫ x2 t (−2tdt) = −2 ∫ (1− t2)2dt = −2 ∫ (1− 2t2 + t4)dt = −2 ( t− 2 3 t3 + 1 5 t5 ) + C = −2 (√ 1− x− 2 3 √ (1− x)3 + 1 5 √ (1− x)5 ) + C. 29. ∫ 2 0 (x− 1)25dx Solution: Let x− 1 = t⇒ dx = dt. Hence∫ (x− 1)25dx = ∫ t25dt = t26 26 + C = (x− 1)26 26 + C. Therefore, ∫ 2 0 (x− 1)25dx = (x− 1) 26 26 ]2 0 = 1 26 − 1 26 = 0. 7 30. ∫ pi 0 sec2(t/4)dt ∫ sec2(t/4)dt = 4 tan(t/4) + C. Therefore,∫ pi 0 sec2(t/4)dt = 4 tan(t/4)]pi0 = 4 tan 0− 4 tanpi/4 = 0− 1 = −1. 31. ∫ 1 0 xe−x 2 dx Solution: Let −x2 = t⇒ dx = −dt 2x . Therefore,∫ xe−x 2 dx = − ∫ xet dt 2x = −1 2 ∫ etdt = −1 2 e−x 2 + C. Consequently,∫ 1 0 xe−x 2 dx = −1 2 e−x 2 ]1 0 = −1 2 e−1 + 1 2 = 1 2 (1− e−1). 32. ∫ 2 −2(x+ 3) √ 4− x2dx Solution:∫ 2 −2 (x+ 3) √ 4− x2dx = ∫ 2 −2 x √ 4− x2dx+ 3 ∫ 2 −2 √ 4− x2dx = −1 3 √ (4− x2)3 ]2 −2 + 3× the area of a semi circle whose radius is 2, where the first integral has been solved using the substitution 4−x2 = t. For the second integral: The equationof the circle centered at the origin with radius 2 is x2 + y2 = 4 ⇒ y = ±√4− x2. Therefore, the area of the upper semicircle is given by the area bounded below by the x−axis and above by the curve y = √ 4− x2. This area may be evaluated by the second integral. The area of this semicircle is 2pi. Standard computations now give: ∫ 2 −2 (x+ 3) √ 4− x2dx = 6pi. 8 33. If f is continuous and ∫ 4 0 f(x)dx = 10, find ∫ 2 0 f(2x)dx. Solution: Do the substitution 2x = t, and observe that the limits of the integral must be changed, in the wanted integral to get∫ 2 0 f(2x)dx = 1 2 ∫ 4 0 f(t)dt = 1 2 × 10 = 5. 34. If f is continuous and ∫ 9 0 f(x)dx = 4, find ∫ 3 0 xf(x2)dx. Solution: Let x2 = t in the wanted integral to get:∫ 3 0 xf(x2)dx = 1 2 ∫ 9 0 f(t)dt = 1 2 × 4 = 2. 35. If a and b are positive numbers, show that∫ 1 0 xa(1− x)bdx = ∫ 1 0 xb(1− x)adx. Solution: Let 1−x = t⇒ dx = −dt. Observe that the limits should be changed as follows: x = 0⇒ t = 1− 0 = 1 and x = 1⇒ t = 1− 1 = 0. Hence∫ 1 0 xa(1− x)bdx = ∫ 0 1 (1− t)atb(−dt) = − ∫ 0 1 (1− t)atbdt = ∫ 1 0 tb(1− t)adt = ∫ 1 0 xb(1− x)adx. Note: When having definite integrals, the letter used for the variable is not a big matter. In other words, ∫ 1 0 tb(1− t)adt = ∫ 1 0 xb(1− x)adx. This observation is not necessarily true when dealing with indefinite in- tegrals! 36. Show that ∫ pi 0 xf(sinx)dx = pi 2 ∫ pi 0 f(sinx)dx. 9 Solution: Let pi − x = t⇒ dx = −dt. Hence∫ pi 0 xf(sinx)dx = ∫ 0 pi (pi − t)f(sin(pi − t))(−dt) = pi ∫ pi 0 f(sin t)dt− ∫ pi 0 tf(sin t)dt, therefore, ∫ pi 0 xf(sinx)dx+ ∫ pi 0 tf(sin t)dt = pi ∫ pi 0 f(sin t)dt ⇒ 2 ∫ pi 0 xf(sinx)dx = pi ∫ pi 0 f(sinx)dx ⇒ ∫ pi 0 xf(sinx)dt = pi 2 ∫ pi 0 f(sinx)dx. 1.2 Integration by parts The rule for integration by parts is∫ udv = uv − ∫ vdu. This suggests that we divide the integrand into two parts. One (u) to be differentiated and the other one (dv) to be integrated. The following are problems taken from section 8.2, Calculus 8th edition, Howard Anton. 1. ∫ xe−2xdx. Whenever we have a product of a polynomial with an expo- nential function we do parts if the exponent of the exponential is linear. If the exponent is not linear we do a substitution. We let the polynomial = u and the exponential = dv. Thus, x e−2x ↓(differentiate) ↘+ ↓(integrate) 1 − −→∫ −1 2 e−2x . 10 In this table: The first column represents the differentiation process and the second represents the integral part. Therefore,∫ xe−2xdx = −1 2 xe−2x − ∫ −1 2 e−2xdx = −1 2 xe−2x − 1 4 e−2x + C. 2. ∫ xe3xdx. Do similar steps: x e3x ↓ ↘+ ↓ 1 − −→∫ 1 3 e3x to get ∫ xe3xdx = 1 3 xe3x − 1 3 ∫ e3xdx = xe3x − 1 9 e3x + C. 3. ∫ x2exdx. We apply the above steps twice as the table indicates: x2 ex ↓ ↘+ ↓ 2x ex ↓ ↘− ↓ 2 −→+∫ ex . ∫ x2exdx = x2ex − 2xex + 2 ∫ exdx = x2ex − 2xex + 2ex + C. 4. ∫ x2e−2xdx. x2 e−2x ↓ ↘+ ↓ 2x −1 2 e−2x ↓ ↘− ↓ 2 −→+∫ 1 4 e−2x . ∫ x2e−2xdx = −1 2 x2e−2x − 1 2 xe−2x + 1 2 ∫ e−2xdx = −1 2 x2e−2x − 1 2 xe−2x − 1 4 e−2x + C. 5. ∫ x sin 3xdx 11 When the integrand is a product of a polynomial and a trigonometric function we let the polynomial = u and the trig. function = dv, provided that the argument of the trig. function is linear. Thus: x sin 3x ↓ ↘+ ↓ 1 −→∫ − −1 3 cos 3x ∫ x sin 3xdx = −1 3 x cos 3x+ 1 3 ∫ cos 3xdx = −1 3 x cos 3x+ 1 9 sin 3x+ C. 6. ∫ x cos 2xdx. Similar steps: x cos 3x ↓ ↘+ ↓ 1 −→∫ − 1 2 sin 2x ∫ x cos 2xdx = 1 2 x sin 2x− 1 2 ∫ sin 2xdx = 1 2 x sin 2x+ 1 4 cos 2x+ C. 7. ∫ x2 cosxdx. We apply integration by parts twice to get: x2 cosx ↓ ↘+ ↓ 2x sin x ↓ ↘− ↓ 2 −→∫ + − cosx ∫ x2 cosxdx = x2 sin x+ 2x cosx− 2 ∫ cosxdx = x2 sin x+ 2x cosx− 2 sin x+ C. 8. ∫ x2 sin xdx. Similar steps: x2 sin x ↓ ↘+ ↓ 2x − cosx ↓ ↘− ↓ 2 −→∫ + − sin x ∫ x2 sin xdx = −x2 cosx+ 2x sin x− 2 ∫ sin xdx = −x2 cosx+ 2x sin x+ 2 cosx+ C. 9. ∫ x lnxdx. 12 When the integrand is a product of a polynomial and a logarithmic function we let the polynomial = dv and the logarithm = u as follows: lnx x ↓ ↘+ ↓ 1 x −→∫ − 1 2 x2 ∫ x lnxdx = 1 2 x2 lnx− 1 2 ∫ xdx = 1 2 x2 lnx− 1 4 x2 + C. 10. ∫ √ x lnxdx. lnx √ x ↓ ↘+ ↓ 1 x −→∫ − 2 3 x3/2 ∫ √ x lnxdx = 2 3 x3/2 lnx− 2 3 ∫ √ xdx = 2 3 x3/2 lnx− 4 9 x3/2 + C. 11. ∫ (lnx)2dx (lnx)2 1 ↓ ↘+ ↓ 2 lnx x −→∫ − x∫ (lnx)2dx = x(lnx)2 − 2 ∫ lnxdx = x(lnx)2 − 2(x lnx− x) + C. We used the fact lnxdx = x lnx − x + C. This can be done using integration by parts. 12. ∫ lnx√ x dx lnx x−1/2 ↓ ↘+ ↓ 1 x −→∫ − 2x1/2∫ lnx√ x dx = 2 √ x lnx− 2 ∫ x−1/2dx = 2 √ x lnx− 4√x+ C. 13 13. ∫ ln(3x−2)dx ln(3x− 2) 1 ↓ ↘+ ↓ 3 3x−2 −→ ∫ − x∫ ln(3x− 2)dx = x ln(3x− 2)− ∫ 3x 3x− 2dx = x ln(3x− 2)− ∫ 3x− 2 + 2 3x− 2 dx = x ln(3x− 2)− ∫ dx− ∫ 2 3x− 2dx = x ln(3x− 2)− x− 2 3 ln(3x− 2) + C. 14. ∫ ln(x2+4)dx ln(x2 + 4) 1 ↓ ↘+ ↓ 2x x2+4 −→∫ − x ∫ ln(x2 + 4)dx = x ln(x2 + 4)− ∫ 2x2 x2 + 4 dx = x ln(x2 + 4)− 2 ∫ ( 1− 4 x2 + 4 ) dx = x ln(x2 + 4)− 2 ( x− arctan(1 2 x) ) + C. 15. ∫ arcsinxdx arcsinx 1 ↓ ↘+ ↓ 1√ 1−x2 −→ ∫ x ∫ arcsinxdx = x arcsinx− ∫ x√ 1− x2dx = x arcsinx+ √ 1− x2 + C, where we have used a substitution in the last integral. 16. ∫ arccos(2x)dx. arccos(2x) 1 ↓ ↘+ ↓ −2√ 1−4x2 −→ ∫ − x ∫ arccos(2x)dx = x arccos(2x) + ∫ 2x√ 1− 4x2dx = x arccos(2x)− 1 2 √ 1− 4x2 + C 17. ∫ arctan(3x)dx arctan(3x) 1 ↓ ↘+ ↓ 3 9x2+1 −→∫ − x 14 ∫ arctan(3x)dx = x arctan(3x)− ∫ 3x 9x2 + 1 dx = x arctan(3x)− 1 6 ln(9x2 + 1) + C. 18. ∫ x arctanxdx arctanx x ↓ ↘+ ↓ 1 x2+1 −→∫ − x2 2 ∫ x arctanxdx = 1 2 x2 arctanx− 1 2 ∫ x2 x2 + 1 dx = 1 2 x2 arctanx− 1 2 ∫ ( 1− 1 x2 + 1 ) dx = 1 2 x2 arctanx− 1 2 (x− arctanx) + C. 19. ∫ ex sin xdx sin x ex ↓ ↘+ ↓ cosx ex ↓ ↘− ↓ − sin x −→∫ + ex ∫ ex sin xdx = ex sin x− ex cosx− ∫ ex sin xdx. Thus, we have ∫ ex sin xdx on both sides. Collect it in one side: 2 ∫ ex sin xdx = ex(sinx− cosx) + C ⇒ ∫ ex sin xdx = 1 2 ex (sinx− cosx) + C. 20. ∫ e3x cos(2x)dx cos(2x) e3x ↓ ↘+ ↓ −2 sin x 1 3 e3x ↓ ↘− ↓ −4 cos(2x) −→∫ + 1 9 e3x ∫ e3x cos(2x)dx = 1 3 e3x cos(2x) + 2 9 e3x sin(2x)− 4 9 ∫ e3x cos(2x)dx. Similar idea to the above: 13 9 ∫ e3x cos(2x)dx = e3x ( 1 3 cos(2x) + 2 9 sin(2x) ) + C ⇒ ∫ e3x cos(2x)dx = 9 13 × e3x ( 1 3 cos(2x) + 2 9 sin(2x) ) + C 15 21. ∫ eax sin(bx)dx sin(bx) eax ↓ ↘+ ↓ b cos(bx) 1 a eax ↓ ↘− ↓ −b2 sin x −→∫ + 1 a2 eax ∫ eax sin(bx)dx = 1 a eax sin(bx)− b a2 eax cos(bx)− b 2 a2 ∫ eax sin(bx)dx; hence, ( 1 + b2 a2 )∫ eax sin(bx)dx = eax ( 1 a sin(bx)− b a2 cos(bx) ) + C ⇒ ∫ eax sin(bx)dx = a2 a2 + b2 eax ( 1 a sin(bx)− b a2 cos(bx) ) + C. 22. ∫ e−3θ sin(5θ)dθ. Apply the above ideas! 23. ∫ sin(lnx)dx. Whenever the argument of a trig. function is not linear we start by substituting this argument. Thus lnx = t⇒ dx = xdt⇒ dx = etdt ∫ sin(lnx) = ∫ sin(t)× etdt = ∫ et sin tdt = 1 2 et (sin t− cos t)+ C = 1 2 elnx (sin(lnx)− cos(lnx)) + C = 1 2 x (sin(lnx)− cos(lnx)) + C. 24. ∫ cos(lnx)dx. Similar to the above! 25. ∫ x sec2 xdx x sec2 x ↓ ↘+ ↓ 1 −→∫ − tan x 16 ∫ x sec2 xdx = x tan x− ∫ tan xdx = x tan x− ln |secx|+ C. Reminder: Recall that∫ tan xdx = ∫ sin x cosx dx may be done by doing the substitution cosx = t. 26. ∫ x tan2 xdx x tan2 x ↓ ↘+ ↓ 1 −→∫ − tan x− x∫ x tan2 xdx = x(tanx− x)− ∫ (tanx− x) dx = x(tanx− x)− ( ln | sec x| − x 2 2 ) + C. 27. ∫ x3ex 2 dx Whenever the exponent of e is not linear, we substitute it! Thus, x2 = t⇒ dx = dt 2x makes∫ x3ex 2 dx = ∫ x3et dt 2x = 1 2 ∫ tetdt = 1 2 et(t− 1) + C = 1 2 ex 2 (x2 − 1) + C. Note: We used integration by parts to do ∫ tetdt. The details are left to the student. 28. ∫ xex (x+1)2 dx. xex (x+ 1)−2 ↓ ↘+ ↓ (x+ 1)ex −→∫ − −(x+ 1)−1∫ xex (x+ 1)2 dx = − xe x x+ 1 + ∫ exdx = − xe x x+ 1 + ex + C = ex x+ 1 + C. 29. ∫ 2 0 xe2xdx. We do ∫ xe2xdx by parts and then we plug the limits: 17 x e2x ↓ ↘+ ↓ 1 −→∫ − 1 2 e2x ∫ xe2xdx = 1 2 xe2x − 1 2 ∫ e2xdx = 1 2 xe2x − 1 4 e2x + C. Therefore, ∫ 2 0 xe2xdx = e2x ( 1 2 x− 1 4 )]2 0 = 3 4 e4 − −1 4 = 1 4 (3e4 + 1). 30. ∫ 1 0 xe−5xdx x e−5x ↓ ↘+ ↓ 1 −→∫ − −1 5 e−5x ∫ xe−5xdx = −1 5 xe−5x − 1 25 e−5x + C, therefore, ∫ 1 0 xe−5xdx = e−5x (−1 5 x− 1 25 )]1 0 = − 6 25 e−5 − −1 25 = 1 25 (1− 6e−5). 31. ∫ e 1 x2 lnxdx lnx x2 ↓ ↘+ ↓ 1 x −→∫ − x3 3 ∫ x2 lnxdx = x3 3 lnx− 1 9 x3 + C, therefore, ∫ e 1 x2 lnxdx = ( 1 3 lnx− 1 9 ) x3 ]e 1 = 2 9 e3 − −1 9 = 1 9 (1 + 2e3). 32. ∫ e√ e lnx x2 dx lnx 1 x2 ↓ ↘+ ↓ 1 x −→∫ − − 1 x ∫ lnx x2 dx = −1 x lnx+ ∫ 1 x2 dx = −1 x lnx− 1 x + C, 18 therefore, ∫ e √ e lnx x2 dx = −1 x (lnx+ 1) ]e √ e = −2 e + 3 2 √ e . 33. ∫ 1 −1 ln(x+2)dx ln(x+ 2) 1 ↓ ↘+ ↓ 1 x+2 −→∫ − x+ 2 ∫ ln(x+ 2)dx = (x+ 2) ln(x+ 2)− x+ C, therefore, ∫ 1 −1 ln(x+ 2)dx = (x+ 2) ln(x+ 2)− x]1−1 = 3 ln 3− 1− 1 = 3 ln 3− 2. 34. ∫ √3/2 0 arcsinxdx. We have seen that∫ arcsinxdx = x arcsinx+ √ 1− x2 + C. For the details see problem 15. Thus,∫ √3/2 0 arcsinxdx = x arcsinx+ √ 1− x2 ]√3/2 0 = √ 3 2 arcsin( √ 3/2) + √ 1/4− (0 + 1) = √ 3pi 6 − 1 2 . 35. ∫ 4 2 sec−1( √ θ)dθ sec−1 √ θ 1 ↓ ↘+ ↓ 1 2θ √ θ−1 −→ ∫ − θ ∫ sec−1 √ θdθ = θ sec−1 √ θ − 1 2 ∫ 1√ θ − 1dθ = θ sec−1 √ θ −√θ − 1 + C. Therefore,∫ 4 2 sec−1( √ θ)dθ = θ sec−1 √ θ −√θ − 1 ]4 2 = 2 sec−1 2− √ 3− 2 sec−1 √ 2 + 1 = 2pi 3 − √ 3− 2pi 4 + 1 = pi 6 + 1− √ 3. 19 36. ∫ 2 1 x sec−1 xdx 37. ∫ pi 0 x sin 2xdx 38. ∫ pi 0 (x+ x cosx)dx 39. ∫ 3 1 √ x tan−1 √ xdx tan−1 √ x √ x ↓ ↘+ ↓ 1 2 √ x(x2+1) −→∫ − 1 2 √ x ∫ √ x tan−1 √ xdx = tan−1 √ x 2 √ x − 1 4 ∫ 1 x(x2 + 1) dx = tan−1 √ x 2 √ x − 1 4 ∫ 1 + x2 − x2 x(x2 + 1) dx = tan−1 √ x 2 √ x − 1 4 ∫ ( 1 x − x x2 + 1 ) dx = tan−1 √ x 2 √ x − 1 4 ( ln |x| − 1 2 ln(x2 + 1) ) + C. Hence,∫ 3 1 √ x tan−1 √ xdx = tan−1 √ x 2 √ x − 1 4 ( ln |x| − 1 2 ln(x2 + 1) )]3 1 = ( tan−1 √ 3 2 √ 3 − 1 4 ( ln 3− 1 2 ln 10 )) − ( tan−1 1 2 − 1 4 × −1 2 ln 2 ) = ...... 40. ∫ 2 0 ln(x2 + 1)dx. 41. (a) ∫ e √ xdx Let √ x = t⇒ x = t2 ⇒ dx = 2tdt. Hence, ∫ e √ xdx = ∫ et2tdt = 2 ∫ tetdt = 2(tet − et) + C = 2e √ x( √ x− 1) + C. (b) ∫ cos √ xdx 20 √ x = t⇒ x = t2 ⇒ dx = 2tdt. ∫ cos √ xdx = 2 ∫ t cos tdt = 2(t sin t+ cos t) + C = 2( √ x sin √ x+ cos √ x) + C 58. In each part, use integration by parts or other methods to derive the reduction formula: (a) ∫ secn xdx = secn−2 x tan x n− 1 + n− 2 n− 1 ∫ secn−1 xdx. secn−2 x sec2 x ↓ ↘+ ↓ (n− 2) secn−2 x tan x −→∫ − tan x ∫ secn xdx = secn−2 x tan x− (n− 2) ∫ secn−2 x tan2 xdx = secn−2 x tan x− (n− 2) ∫ secn−2 x(sec2 x− 1)dx = secn−2 x tan x− (n− 2) ∫ secn xdx+ (n− 2) ∫ secn−2 xdx, hence ∫ secn xdx+ (n− 2) ∫ secn xdx = secn−2 x tan x+ (n− 2) ∫ secn−2 xdx ⇒ (n− 1) ∫ secn xdx = secn−2 x tan x+ (n− 2) ∫ secn−2 xdx ⇒ ∫ secn xdx = secn−2 x tan x n− 1 + n− 2 n− 1 ∫ secn−2 xdx. (b) ∫ tann xdx = tann−1 x n− 1 − ∫ tann−2 xdx. 21 ∫ tann xdx = ∫ tann−2 x tan2 xdx = ∫ tann−2 x(sec2 x− 1)dx = ∫ tann−2 x sec2 xdx− ∫ tann−2 xdx = tann−1 x n− 1 − ∫ tann−2 xdx, where we have used the substitution tanx = t to do the integral∫ tann−2 x sec2 xdx. (c) ∫ xnexdx = xnex − n ∫ xn−1exdx. Integrate by parts: xn ex ↓ ↘+ ↓ nxn−1 −→∫ − ex∫ xnexdx = xnex − n ∫ xn−1exdx. 1.3 Trigonometric Integrals 1. ∫ cos3 x sin xdx. Since the derivative of cos x is sin x which is a part of the integrand we do the substitution: cosx = t⇒ − sin xdx = dt⇒ dx = − dt sinx . Thus, ∫ cos3 x sin xdx = ∫ t3 sin x −dt sin x = − ∫ t3dt = −t 4 4 + C = −cos 4 x 4 + C. 2. ∫ sin5 3x cos 3xdx. 22 sin 3x = t⇒ 3 cos 3xdx = dt⇒ dx = dt 3 cos 3x . Hence ∫ sin5 3x cos 3xdx = ∫ t5 cos 3x dt 3 cos 3x dx = 1 3 ∫ t5dt = 1 18 t18 + C = 1 18 sin6 3x+ C. 3. ∫ sin2 5xdx. To do the integral ∫ sineven xdx we use the identity sin2 x = 1 2 (1− cos 2x). Thus, ∫ sin2 5xdx = 1 2 ∫ (1− cos 10x)dx = 1 2 ( x− sin 10x 10 ) + C. 4. ∫ cos2 3xdx. To do ∫ coseven xdx we use the identity cos2 x = 1 2 (1 + cos 2x). Thus, ∫ cos2 3xdx = 1 2 ∫ (1 + cos 6x)dx = 1 2 ( x+ sin 6x 6 ) + C. 5. ∫ sin3 axdx. To do ∫ sinodd xdx we let cosx = t⇒ dx = − dt sinx and then we use the identity sin2 x = 1− cos2 x = 1− t2. 23 Let cos ax = t⇒ dx = − dt a sin ax (a 6= 0),∫ sin3 axdx = ∫ sin3 ax −dt a sin ax = −1 a ∫ sin2 axdt = −1 a ∫ (1− cos2 ax)dt = −1 a ∫ (1− t2)dt = −1 a (t− t3/3) + C = −1 a ( cos ax− cos 3 ax 3 ) + C. 6. ∫ cos3 atdt. Let sin at = x⇒ dt = dx a cos at , hence∫ cos3 atdt = ∫ cos3 at dx a cos at = 1 a ∫ cos2 atdx = 1 a ∫ (1− sin2 at)dx = 1 a ∫ (1− x2)dx = 1 a (x− x3/3) + C = 1 a ( sin at− sin 3 at 3 ) + C. 7. ∫ sin ax cos axdx. Let sin ax = t ⇒ a cos axdx = dt ⇒ dx = 1 a cos ax dt, hence ∫ sin ax cos axdx = ∫ t cos ax 1 a cos ax dt = 1 a ∫ tdt = 1 2a t2 + C = 1 2a sin2 ax+ C. 8. ∫ sin3 x cos3 xdx. To do ∫ sinodd x cosodd xdx we let the trig. function whose power is bigger equal t. Thus, let sinx = t⇒ cosxdx = dt⇒ dx = dt cosx , hence∫ sin3 x cos3 xdx = ∫ t3 cos3 x dt cosx = ∫ t3 cos2 xdt = ∫ t3(1− t2)dt = ∫ (t3 − t5)dt = ( t4 4 − t 6 6 ) + C = ( sin4 x 4 − sin 6 x 6 ) + C. 24 9. ∫ sin2 t cos3 tdt. To do ∫ sineven x cosodd xdx we let sinx = t, then we use cos2 x = 1− sin2 x = 1− t2. Thus,∫ sin2 x cos3 xdx = ∫ t2 cos3 x dt cosx = ∫ t2 cos2 xdt = ∫ t2(1− sin2 x)dt = ∫ t2(1− t2)dt = t3/3− t5/5 + C = sin 3 x 3 − sin 5 x 5 + C. 10. ∫ sin3 x cos2 xdx To do ∫ sinodd x coseven xdx we let cosx = t⇒ dx = − 1 sinx dtthen we use sin2 x = 1− cos2 x = 1− t2. Thus ∫ sin3 x cos2 xdx = ∫ sin3 xt2 −dt sin x = − ∫ sin2 xt2dt = − ∫ (1− t2)t2dt = − ( t3 3 − t 5 5 ) + C = − ( cos3 x 3 − cos 5 x 5 ) + C. 11. ∫ sin2 x cos2 xdx. To do ∫ sineven x coseven xdx, we use the identities sin2 x = 1 2 (1− cos 2x), cos2 x = 1 2 (1 + cos 2x). Thus, ∫ sin2 x cos2 xdx = ∫ 1 2 (1− cos 2x)1 2 (1 + cos 2x)dx = 1 4 ∫ (1− cos2 2x)dx = 1 4 ∫ sin2 2xdx = 1 4 ∫ 1 2 (1− cos 4x)dx = 1 8 ( x− sin 4x 4 ) + C. 25 12. ∫ sin2 x cos4 xdx. Apply the above remark,∫ sin2 x cos4 xdx = ∫ 1 2 (1− cos 2x) [ 1 2 (1 + cos 2x) ]2 dx = 1 8 ∫ (1− cos 2x)(1 + cos 2x)(1 + cos 2x)dx = 1 8 ∫ (1− cos2 2x)(1 + cos 2x)dx = 1 8 ∫ sin2 2x(1 + cos 2x)dx = 1 8 ∫ (sin2 2x+ sin2 2x cos 2x)dx. Now, for∫ sin2 2xdx = 1 2 ∫ (1− cos 4x)dx = 1 2 ( x− sin 4x 4 ) + C. For ∫ sin2 2x cos 2xdx we let sin 2x = t to get∫ sin2 2x cos 2xdx = 1 6 sin3 2x+ C. Thus ∫ sin2 x cos4 xdx = 1 8 [ 1 2 ( x− sin 4x 4 ) + 1 6 sin3 2x ] + C. 13. ∫ sin 2x cos 3xdx. To do integrals of the form ∫ sin ax cos bxdx we use the identity sin ax cos bx = 1 2 [sin(a+ b)x+ sin(a− b)x] . Thus ∫ sin 2x cos 3xdx = 1 2 ∫ [sin 5x+ sin(−x)] dx = 1 2 [ cos 5x 5 + cosx ] + C. 14. ∫ sin 3x cos 2xdx. We do similar steps:∫ sin 3x cos 2xdx = 1 2 [sin 5x+ sinx] dx = 1 2 [− cos 5x 5 − cosx ] + C. 26 15. ∫ sin x cos(x/2)dx. Apply the above identity:∫ sin x cos(x/2)dx = 1 2 ∫ [sin(3x/2) + sin(x/2)] dx = 1 2 [ −cos(3x/2) 3/2 − cos(x/2) 1/2 ] + C. 16. ∫ cos1/3 x sin xdx. Do the substitution: cosx = t⇒ dx = − dt sinx Thus ∫ cos1/3 x sinxdx = ∫ t1/3 sin x −dt sin x = −3 4 t4/3 + C = −3 4 cos4/3 x+ C. 17. ∫ pi/2 0 cos3 xdx. To do ∫ cosodd xdx we let sin x = t⇒ dx = dt cosx . Therefore,∫ cos3 xdx = ∫ cos3 x dt cosx = ∫ cos2 xdt = ∫ (1− t2)dt = t− t 3 3 + C = sinx− sin 3 x 3 + C. Hence, ∫ pi/2 0 cos3 xdx = sin x− sin 3 x 3 ]pi/2 0 = 2 3 − 0 = 2 3 . 27 18. ∫ pi/2 0 sin2 x 2 cos2 x 2 dx = ∫ pi/2 0 1 2 (1− cosx)1 2 (1 + cos x)dx = 1 4 ∫ pi/2 0 (1− cos2 x)dx = 1 4 ∫ pi/2 0 sin2 xdx = 1 4 1 2 ∫ (1− cos 2x)dx = 1 8 ( x− sin 2x 2 )]pi/2 0 = pi 16 . 19. ∫ pi/3 0 sin4 3x cos3 3xdx. We do the substitution sin 3x = t⇒ dx = dt 3 cosx . ∫ sin4 3x cos3 3xdx = ∫ t4 cos3 3x dt 3 cos 3x = 1 3 ∫ t4 cos2 3xdx = 1 3 ∫ t4(1− t2)dt = 1 3 ( t5 5 − t 7 7 ) + C = 1 3 ( sin5 3x 5 − sin 7 3x 7 ) + C. Therefore,∫ pi/3 0 sin4 3x cos3 3xdx = 1 3 ( sin5 3x 5 − sin 7 3x 7 )]pi/3 0 = 0. 20. ∫ pi −pi cos 2 5xdx. ∫ cos2 5xdx = 1 2 ∫ (1 + cos 10x)dx = 1 2 ( x+ sin 10x 10 ) + C. Hence ∫ pi −pi cos2 5xdx = 1 2 ( x+ sin 10x 10 )]pi −pi = pi. 28 21. ∫ pi/6 0 sin 4x cos 2xdx.∫ sin 4x cos 2xdx = 1 2 ∫ [sin 6x+ sin 2x] dx = 1 2 [− cos 6x 6 − sin 2x 2 ] + C. Thus ∫ pi/6 0 sin 4x cos 2xdx = 1 2 [− cos 6x 6 − sin 2x 2 ]pi/6 0 = 1 2 [ 1 6 − √ 3 4 ] − 1 2 [−1 6 − 0] = 1 6 − √ 3 4 . 22. ∫ 2pi 0 sin2 kxdx. ∫ sin2 kxdx = 1 2 ∫ (1− cos 2kx)dx = 1 2 ( x− sin 2kx 2k ) + C. Hence, ∫ 2pi 0 sin2 kxdx = 1 2 ( x− sin 2kx 2k )]2pi 0 = pi. 23. ∫ sec2(2x − 1)dx. Doing the substitution 2x − 1 = t yields ∫ sec2(2x − 1)dx = 1 2 tan(2x− 1) + C. 24. ∫ tan 5xdx = ∫ sin 5x cos 5x dx. doing the substitution cos 5x = t yields 1 5 ∫ tan 5xdx = ln | sec 5x|+ C. 25. ∫ e−x tan(e−x)dx. Do the substitution e−x = t to get∫ e−x tan(e−x)dx = − ∫ tan tdt = − ln | sec t|+ C = − ln | sec(e−x)|+ C. 29 26. ∫ cot 3xdx = ∫ cos 3x sin 3x dx. Do the substitution cos 3x = t to get∫ cot 3xdx = ∫ cos 3x t dt 3 cos 3x = 1 3 ln |t|+ C = 1 3 ln | sin 3x|+ C. 27. ∫ sec 4xdx. We multiply with and divide by sec 4x+ tan 4x to get∫ sec 4xdx = ∫ sec 4x(sec 4x+ tan 4x) sec 4x+ tan 4x dx = ∫ sec2 4x+ sec 4x tan 4x tan 4x+ sec 4x dx = 1 4 ln | sec 4x+ tan 4x|+ C, where we have used the substitution sec 4x + tan 4x = t in the last integral. 28. ∫ sec √ x√ x dx. Do the substitution √ x = t⇒ x = t2 ⇒ dx = 2tdt. Therefore∫ sec √ x√ x dx = 2 ∫ sec tdt = 2 ∫ sect(sec t+ tan t) sec t+ tan t dt = 2 ln | sec t+ tan t|+ C = 2 ln | sec√x+ tan√x|+ C. 29. ∫ tan2 x sec2 xdx. Do the substitution: tan x = t⇒ sec2 xdx = dt⇒ dx = dt sec2 x . Hence ∫ tan2 x sec2 xdx = ∫ t2 sec2 x dt sec2 x = ∫ t2dt = t3 3 + C = tan3 x 3 + C. 30. ∫ tan5 x sec4 xdx. 30 To do integrals of the form ∫ tanodd x seceven xdx, do one of the substitutions tan x = t or sec x = t. Let us do tanx = t⇒ sec2 xdx = dt⇒ dx = dt sec2 x . Hence∫ tan5 x sec4 xdx = ∫ t5 sec4 x dt sec2 x = ∫ t5 sec2 xdt = ∫ t5(tan2 x+ 1)dt = ∫ t5(t2 + 1)dt = t8 8 + t6 6 + C = tan8 x 8 + tan6 x 6 + C. 31. ∫ tan 4x sec4 4xdx. Similar to the above. Let’s do sec 4x = t⇒ dx = dt 4 sec 4x tan 4x = dt 4t tan 4x . Hence, ∫ tan 4x sec4 4xdx = 1 4 ∫ tan 4xt4 dt t tan 4x = 1 4 ∫ t3dt = 1 4 t4 4 + C = sec4 4x 16 + C. 32. ∫ tan4 x sec4 xdx. To do ∫ taneven x seceven xdx we let tan x = t, then we use the identity sec2 x = tan2 x+ 1. Thus, ∫ tan4 x sec4 xdx = ∫ t4 sec4 x dt sec2 x = ∫ t4 sec2 xdt = ∫ t4(tan2 x+ 1)dt = ∫ t4(t2 + 1)dt = t7 7 + t5 5 + C = tan7 x 7 + tan5 x 5 + C. 33. ∫ sec5 x tan3 xdx. 31 To do integrals of the form ∫ tanodd x secodd xdx we use the substitution sec x = t then we use the identity tan2 x = sec2 x− 1. Whence∫ sec5 x tan3 xdx = ∫ t5 tan3 x dt sec x tan x = ∫ t5 tan3 x dt t× tan x = ∫ t4 tan2 xdt = ∫ t4(sec2 x− 1)dt = ∫ t4(t2 − 1)dt = t 7 7 − t 5 5 + C = sec7 x 7 − sec 5 x 5 + C. 34. tan5 x sec xdx. Same as above: secx = t ⇒ secx tan xdx = dt ⇒ dx = dt t×tanx . Hence∫ tan5 x sec xdx = ∫ tan5 xt dt t× tan x = ∫ tan4 xdt = ∫ (tan2 x)2dt = ∫ (sec2 x− 1)2dt = ∫ (t2 − 1)2dt = ∫ (t4 − 2t2 + 1)dt = t 5 5 − 2t 3 3 + t+ C = sec5 x 5 − 2 sec 3 x 3 + sec x+ C. 35. ∫ tan4 x sec xdx. For integrals of the form ∫ taneven x secodd xdx we replace each tan2 x with sec2 x− 1 in order to get an integral involving only powers of sec x. Thus∫ tan4 x sec xdx = ∫ (tan2 x)2 sec xdx = ∫ (sec2 x− 1)2 sec xdx = ∫ (sec4 x− 2 sec2 x+ 1) sec xdx = ∫ (sec5 x− 2 sec3 x+ sec x)dx. (1.3.1) 32 Each of these three integrals has the form ∫ secodd xdx. For ∫ sec xdx see example 27 above. Thus∫ sec xdx = ln | sec x+ tanx|+ C. Now, To do integrals of the form ∫ secodd>1 xdx we use integration by parts. sec2 x = dv and the rest is u. Thus, for ∫ sec3 xdx we use sec2 x = dv, sec x = u: sec x sec2 x ↓ ↘+ ↓ sec x tan x −→∫ − tan x Therefore∫ sec3 xdx = sec x tan x− ∫ sec x tan2 xdx = sec x tan x− ∫ sec x(sec2 x− 1)dx = sec x tan x− ∫ sec3 xdx+ ∫ sec xdx thus, the term ∫ sec3 x appears on both sides. Collect it to get 2 ∫ sec3 xdx= secx tan x+ ∫ sec xdx ⇒ ∫ sec3 xdx = 1 2 (sec x tan x+ ln | sec x+ tanx|) + C. It remains to do ∫ sec5 xdx. This may be done in a similar way: sec3 x sec2 x ↓ ↘+ ↓ 3 sec3 x tanx −→∫ − tanx ∫ sec5 xdx = sec3 x tanx− 3 ∫ sec3 x tan2 xdx = sec3 x tanx− 3 ∫ sec3 x(sec2 x− 1)dx where we have used the idea for doing integrals of the form ∫ secodd x taneven xdx in the last line. Thus∫ sec5 xdx = sec3 x tan x− 3 ∫ sec5 xdx+ 3 ∫ sec3 xdx. 33 We have ∫ sec5 xdx on both sides, collect it. Also we have ∫ sec3 xdx which has been done above: 4 ∫ sec5 xdx = sec3 x tan x+ 3 2 (sec x tan x+ ln | sec x+ tanx|) + C ⇒ ∫ sec5 xdx = 1 4 [ sec3 x tan x+ 3 2 (sec x tan x+ ln | sec x+ tanx|) ] + C. Thus, collecting the terms needed to do 1.3.1 we see that:∫ tan4 x sec xdx = 1 4 [ sec3 x tan x+ 3 2 (sec x tan x+ ln | sec x+ tan x|) ] − (sec x tan x+ ln | sec x+ tan x|) + ln | sec x+ tanx|+ C. 36. ∫ tan2 x sec3 xdx. Since the power of tan is even and that of sec is odd we may apply the remark above to get∫ tan2 x sec3 xdx = ∫ (sec2 x− 1) sec3 xdx = ∫ sec5 xdx− ∫ sec3 xdx = 1 4 [ sec3 x tan x+ 3 2 (sec x tan x+ ln | sec x+ tanx|) ] − 1 2 (sec x tan x+ ln | sec x+ tan x|) + C, where we have used the formulae of ∫ sec5 xdx and ∫ sec3 xdx made in the above example. 37. ∫ tan t sec3 tdt. Since both powers are odd we let sec t = x⇒ sec t tan tdt = dx⇒ dt = dx x tan t . Thus∫ tan t sec3 tdt = ∫ tan tx3 dx x× tan t = ∫ x2dx = x3 3 + C = sec3 t 3 + C. 38. tan x sec5 xdx. Same as above:∫ tan x sec5 xdx = ∫ tan xt5 dt t× tan x = ∫ t4dt = t5 5 + C = sec5 x 5 + C. 34 39. ∫ sec4 xdx. To do integrals of the form ∫ seceven xdx we let tan x = t. tan x = t⇒ dx = dt sec2 x and sec2 x = t2 + 1. Therefore∫ sec4 xdx = ∫ sec4 x dt sec2 x = ∫ sec2 xdt = ∫ (t2 + 1)dt = t3 3 + t+ C = tan3 x 3 + tanx+ C. 40. ∫ sec5 xdx. This has been done in example 35. 41. ∫ tan3 4xdx. For integrals of the form ∫ tanany power>1 xdx we separate a tan2 xdx and replace it by sec2 x− 1. We repeat this until the problem is solved! ∫ tan3 4xdx = ∫ tan 4x(tan2 4x)dx = ∫ tan 4x(sec2 4x− 1)dx = ∫ tan 4x sec2 4xdx− ∫ tan 4xdx = 1 4 tan2 4x− 1 4 ln | sec 4x|+ C where we have used the substitutions tan 4x = t and cos 4x = t, respec- tively, to do the last two integrals. 42. ∫ tan4 xdx. Same idea as above:∫ tan4 xdx = ∫ tan2 x tan2 xdx = ∫ tan2 x(sec2 x− 1)dx = ∫ tan2 x sec2 xdx− ∫ tan2 xdx = ∫ tan2 x sec2 xdx− ∫ (sec2 x− 1)dx = 1 3 tan3 x− (tanx− x) + C. 35 43. ∫ √ tan x sec4 xdx. Do the substitution tan x = t⇒ sec2 xdx = dt⇒ dx = dt sec2 x . Hence∫ √ tan x sec4 xdx = ∫ √ t sec4 x dt sec2 x = ∫ √ t sec2 xdt = ∫ √ t(tan2 x+ 1)dt = ∫ √ t(t2 + 1)dt = 2 7 t7/2 + 2 3 t3/2 + C = 2 7 tan7/2 x+ 2 3 tan3/2 x+ C. 44. ∫ tan x sec3/2 xdx. Do the substitution sec x = t⇒ sec x tan xdx = dt⇒ dt t tan x hence∫ tan x sec3/2 xdx = ∫ tan xt3/2 dt t tan x = ∫ √ tdt = 2 3 t3/2 + C = 2 3 sec3/2 x+ C. 45. ∫ pi/8 0 tan2 2xdx = ∫ pi/8 0 (sec2 2x− 1)dx = 1 2 tan 2x− x ]pi/8 0 = 1 2 − pi 8 . 46. ∫ pi/6 0 sec3 2x tan 2xdx = 1 6 sec3 2x ]pi/6 0 = 7 6 . 36 47. ∫ pi/2 0 tan5 x 2 dx.∫ tan5 xdx = ∫ tan3 x(sec2 x− 1)dx = ∫ tan3 x sec2 xdx− ∫ tan3 xdx = 1 4 tan4 x− ∫ tan x(sec2 x− 1)dx = 1 4 tan4 x− 1 2 tan2 x+ ln | sec x|+ C. Hence∫ tan5 x 2 dx = 1 2 [ 1 4 tan4 x 2 − 1 2 tan2 x 2 + ln | sec x 2 ] + C, which gives∫ pi/2 0 tan5 x 2 dx = 1 2 [ 1 4 tan4 x 2 − 1 2 tan2 x 2 + ln | sec x 2 | ]pi/2 0 = 1 2 (−1 4 + ln √ 2). 48. ∫ 1/4 0 sec pix tan pixdx = 1 pi sec pix ]1/4 0 = 1 pi (1− √ 2). 49. ∫ cot3 x csc3 xdx. Do the substitution csc x = t⇒ − csc x cotxdx = dt⇒ dx = dt−t cotx. Hence∫ csc3 x cot3 xdx = ∫ t3 cot3 x dt −t cotx = − ∫ t2 cot2 xdt = − ∫ t2(t2 − 1)dt = − ( t5 5 − t 3 3 ) + C = − ( csc5 x 5 − csc 3 x 3 ) + C. 37 50. ∫ cot2 3x sec 3xdx = ∫ cos2 3x sin2 3x 1 cos 3x dx = ∫ cos 3x sin2 3x dx = ∫ cot 3x csc 3xdx = −1 3 csc 3x+ C. 51. ∫ cot3 xdx = ∫ cotx cot2 xdx = ∫ cotx(csc2 x− 1)dx = ∫ cotx csc2 xdx− ∫ cotxdx = −cot 2 x 2 − ln | sin x|+ C. 52. ∫ csc4 xdx. Lett cotx = t⇒ − csc2 xdx = dt⇒ dx = − dt csc2 x . Hence ∫ csc4 xdx = ∫ csc4 x −dt csc2 x = − ∫ csc2 xdt = − ∫ (t2 + 1)dt = − ( t3 3 + t ) + C = − ( cot3 x 3 + cotx ) + C. 53. Let m and n be distinct integers, show that (a) ∫ 2pi 0 sinmx cosnxdx = 0. Use the identity sin a cos b = 1 2 [sin(a− b) + sin(a+ b)]. 38 ∫ 2pi 0 sinmx cosnxdx = 1 2 ∫ 2pi 0 [sin(m− n)x+ sin(m+ n)x] dx = 1 2 [ −cos(m− n)x m− n − cos(m+ n)x m+ n ]2pi 0 = 1 2 [ − 1 m− n − 1 m+ n ] − 1 2 [ −cos(m− n)2pi m− n − cos(m+ n)2pi m+ n ] = 1 2 [ − 1 m− n − 1 m+ n ] − 1 2 [ − 1 m− n − 1 m+ n ] = 0. (b) ∫ cosmx cosnxdx = 0. Use the identity cos a cos b = 1 2 [cos(a− b) + cos(a+ b)] and apply similar ideas as above. (c) ∫ sinmx sinnxdx = 0. Use the identity sin a sin b = 1 2 [cos(a− b)− cos(a+ b)] and apply similar ideas as above. 54. Solve ∫ sinnx cosmxdx when n = m 6= 0. Let’s denote m and n both by a so that we have∫ sinnx cosmxdx = ∫ sin ax cos axdx = 1 2a sin2 ax+ C, where we have used the substitution sin ax = t to do the last integral. 39 1.4 Trigonometric substitution The idea of this section is to compare some algebraic quantities with one of the following trigonometric identities: 1− sin2 t = cos2 t, 1 + tan2 t = sec2 t, sec2 t− 1 = tan2 t. We shall use the symbol ∼ to say that two quantities look like each other. For instance, 1 − x2 ∼ 1 − sin2 t, 2 − 3x2 ∼ 1 − sin2 t, x2 + 3 ∼ tan2 t + 1 and x2 − 3 ∼ sec2 t− 1. The examples illustrate how to choose the suitable substitution once a similar term has been found. The following remark should be emphasized before proceeding: We know that √ cos2 t = | cos t|. In the sequel, we shall always write √cos2 t = cos t. The reason for this is that when we do a substitution like x = sin t we intend to have a well defined inverse sin−1 t of sin t. In order to have this, t should be in [−pi 2 , pi 2 ] where cos t is always positive! 1. ∫ √ 4− x2dx. We observe that 4− x2 ∼ 1− sin2 t⇒ 4− x2 ∼ 4− 4 sin2 t⇒ −x2 ∼ −4 sin2 t⇒ x ∼ 2 sin t. Thus, do the substitution x = 2 sin t⇒ dx = 2 cos tdt. Therefore,∫ √ 4− x2dx = ∫ √ 4− 4 sin2 t× 2 cos tdt = 4 ∫ √ cos2 t cos tdt = 4 ∫ cos2 tdt = 4 ∫ 1 2 (1 + cos 2t)dt = 2 ( t+ 1 2 sin 2t ) + C. 40 In order to find this final answer in terms of x we observe that x = 2 sin t⇒ x 2 = sin t⇒ t = sin−1 x 2 and that sin 2t = 2 sin t cos t⇒ sin 2t = 2x 2 √ 1− sin2 t = x √ 1− x 2 4 . Thus, ∫ √ 4− x2dx = 2 ( sin−1 x 2 + 1 2 x √ 1− x 2 4 ) + C. 2. ∫ √ 1− 4x2dx. We observe that 1− 4x2 ∼ 1− sin2 t⇒ −4x2 ∼ sin2 t⇒ x2 ∼ 1 4 sin2 t⇒ x ∼ 1 2 sin t. Thus, do the substitution x = 1 2 sin t⇒ dx = 1 2 cos tdt. Therefore,∫ √ 1− 4x2dx = ∫ √ 1− 41 4 sin2 t× 1 2 cos tdt = 1 2 ∫ cos2 tdt = 1 4∫ (1 + cos 2t)dt = 1 4 ( t+ 1 2 sin 2t ) + C. In order to write this final answer in terms of x, we observe that x = 1 2 sin t⇒ t = sin−1 2x and sin 2t = 2 sin t cos t = 2× 2x √ 1− sin2 t = 4x √ 1− 4x2. Thus ∫ √ 1− 4x2dx = 1 4 ( sin−1 2x+ 2x √ 1− 4x2 ) + C. 3. ∫ x2√ 16−x2dx. 41 16− x2 ∼ 1− sin2 t⇒ 16− x2 ∼ 16− 16 sin2 t⇒ x ∼ 4 sin t. Do the substitution x = 4 sin t⇒ dx = 4 cos tdt to get ∫ x2√ 16− x2dx = ∫ 16 sin2 t√ 16− 16 sin2 t 4 cos tdt = 16 ∫ sin2 tdt = 8 ∫ (1− cos 2t)dt = 8 ( t− 1 2 sin 2t ) + C = 8 ( sin−1 x 4 − sin t cos t ) + C = 8 ( sin−1 x 4 − x 4 √ 1− x 2 16 ) + C. 4. ∫ dx x2 √ 9−x2 . 9− x2 ∼ 1− sin2 t⇒ 9− x2 ∼ 9− 9 sin2 t⇒ x2 ∼ 9 sin2 t. Thus do the substitution x = 3 sin t⇒ dx = 3 cos tdt. Hence ∫ dx x2 √ 9− x2 = ∫ 3 cos tdt 9 sin2 t √ 9− 9 sin2 t = 1 9 ∫ 1 sin2 t dt = 1 9 ∫ csc2 tdt = −1 9 cot t+ C = −1 9 cos t sin t + C = −1 9 √ 1− x2/9 x/3 + C. 5. ∫ dx (x2+4)2 . x2 + 4 ∼ tan2 t+ 1⇒ x2 + 4 ∼ 4 tan2 t+ 4⇒ x2 ∼ 4 tan2 t. Thus, do the substitution x = 2 tan t⇒ dx = 2 sec2 tdt. 42 Whence ∫ dx (x2 + 4)2 = ∫ 2 sec2 tdt (4 tan2 t+ 4)2 = 1 8 ∫ sec2 t (tan2 t+ 1)2 = 1 8 ∫ 1 sec2 t dt = 1 8 ∫ cos2 tdt = 1 16 (t+ 1 2 sin 2t) + C = 1 16 ( tan−1 x 2 + 2x x2 + 4 ) + C. 6. ∫ x2√ 5+x2 dx. 5 + x2 ∼ 1 + tan2 t⇒ 5 + x2 ∼ 5 + 5 tan2 t⇒ x ∼ 5 tan2 t. Do the substitution x = √ 5 tan t⇒ dx = √5 sec2 tdt to get ∫ x2√ 5 + x2 dx = ∫ 5 tan2 t√ 5 + 5 tan2 t √ 5 sec2 tdt = 5 ∫ tan2 t sec2 t√ 1 + tan2 t dt = 5 ∫ tan2 t sec tdt = 5 ∫ (sec2 t− 1) sec tdt = 5 ∫ (sec3− sec t)dt = 5 ( 1 2 (sec t tan t− ln | sec t+ tan t|)− ln | sec t+ tan t| ) + C = 5 2 sec t tan t− 7 2 ln | sec t+ tan t|+ C = 5 2 √ 1 + x2 5 x√ 5 − 7 2 ln ∣∣∣∣∣ √ 1 + x2 5 + x√ 5 ∣∣∣∣∣+ C. 7. ∫ √ x2−9 x dx. Do the substitution x = 3 sec t⇒ dx = 3 sec t tan tdt. Hence∫ √ x2 − 9 x dx = ∫ 3 tan t 3 sec t 3 sec t tan tdt = 3 ∫ tan2 tdt = 3 ∫ (sec2 t− 1)dt = 3(tan t− t) + C = 3 (√ x2 9 − 1− sec−1 x 3 ) + C. 43 8. ∫ dx x2 √ x2−16 . let x = 4 sec t⇒ dx = 4 sec t tan tdt. Hence∫ dx x2 √ x2 − 16 = ∫ 4 sec t tan t 16 sec2 t √ 16 sec2 t− 16 = 1 16 ∫ 1 sec t dt = 1 16 ∫ cos t+ C = 1 16 sin t+ C = 1 16 √ 1− 16 x2 + C. 9. ∫ 3x3√ 1−x2dx. We may use the substitution 1− x2 = t⇒ dx = − dt2x to get∫ 3x3√ 1− x2dx = ∫ 3x3√ t −dt 2x = −3 2 ∫ x2√ t dt = −3 2 ∫ 1− t√ t dt = −3 2 ( 1√ t −√t ) dt = −3 2 ( 2 √ t− 2 3 t3/2 ) + C = −3 2 ( 2 √ 1− x2 − 2 3 √ (1− x2)3 ) + C. 10. ∫ x3 √ 5− x2dx. Do 5− x2 = t⇒ dx = − dt 2x , hence∫ x3 √ 5− x2dx = − ∫ x3 √ t dt 2x = −1 2 ∫ x2 √ tdt = −1 2 ∫ x2 √ tdt = −1 2 ∫ (5− t)√tdt = −1 2 ∫ (5t1/2 − t3/2)dt = −1 2 ( 10 3 t3/2 − 2 5 t5/2 ) + C = −1 2 ( 10 3 √ (5− x2)3 − 2 5 √ (5− x2)5 ) + C. 11. ∫ dx x2 √ 9x2−4 . Let x = 2 3 sec t⇒ dx = 2 3 sec t tan tdt. 44 Hence ∫ dx x2 √ 9x2 − 4 = ∫ 2/3 sec t tan tdt 4/9 sec2 t √ 4 sec2 t− 4 = 3 4 ∫ 1 sec t dt = 3 4 ∫ sec tdt = 3 4 ln | sec t+ tan t|+ C = 3 4 ln ∣∣∣∣∣3x2 + √ 9x2 4 − 1 ∣∣∣∣∣+ C. 12. ∫ √ 1+t2 t dt. Let t = tan θ ⇒ dt = sec2 θdθ. Thus∫ √ 1 + t2 t dt = ∫ sec θ tan θ sec2 θdθ = ∫ 1/ cos θ sin θ/ cos θ 1 cos2 θ dθ = ∫ 1 sin θ cos2 θ dθ = ∫ sin2 θ + cos2 θ sin θ cos2 θ dθ = ∫ ( sin θ cos2 θ + 1 sin θ ) + C = 1 cos θ + ln | csc θ − cot θ|+ C = √ 1 + t2 + ln ∣∣∣√1 + 1/t2 − 1/t∣∣∣+ C. 13. ∫ 1 (1−x2)3/2dx. Let x = sin t⇒ dx = cos tdt. Whence ∫ 1 (1− x2)3/2dx = ∫ 1 cos3 t cos tdt = ∫ sec2 tdt = tan t+ C = x√ 1− x2 + C. 14. ∫ dx x2 √ x2+25 . Let x = 5 tan t⇒ dx = 5 sec2 tdt. 45 Hence ∫ dx x2 √ x2 + 25 = ∫ 5 sec2 tdt 25 tan2 t √ 25 + 25 tan2 t = 1 25 ∫ sec t tan2 t dt = 1 25 ∫ cos t sin2 t dt = 1 25 ∫ csc t cot tdt = − 1 25 csc t+ C = − 1 25 √ 1 + x2 25 + C. 15. ∫ dx√ x2−9 . Let x = 3 sec t⇒ dx = 3 sec t tan tdt. Hence ∫ dx√ x2 − 9 = ∫ 3 sec t tan t 3 tan t dt = ∫ sec tdt = ln | sec t+ tan t|+ C = ln ∣∣∣∣∣x3 + √ x2 9 − 1 ∣∣∣∣∣+ C. 16. ∫ dx 1+2x2+x4 . By letting x = tan t we see that∫ dx 1 + 2x2 + x4 = ∫ dx (x2 + 1)2 = ∫ sec2 t sec4 t dt = ∫ cos2 tdt = 1 2 ∫ (1 + cos 2t)dt = 1 2 ( t+ sin 2t 2 ) + C = 1 2 ( tan−1 x+ x x2 + 1 ) + C. 17. ∫ dx (4x2−9)3/2 . Let x = 3 sec t 2 ⇒ dx = 3 2 sec t tan tdt. ∫ dx (4x2 − 9)3/2 = ∫ 3/2 sec t tan tdt 27 tan3 t = 1 18 ∫ sec t tan2 t dt = 1 18 ∫ cot t csc tdt = − 1 18 csc t+ C = − 1 18 2x√ 4x2 − 9 + C. 46 18. ∫ 3x3√ 25−x2dx. Do the substitution 25− x2 = t⇒ dx = − dt 2x and x2 = 25− t. We leave the details to the student. 19. ∫ ex √ 1− e2xdx. Use the substitution ex = sin t⇒ dx = cos tdt ex to get ∫ ex √ 1− e2xdx = ∫ ex √ 1− sin2 tcos tdt ex = ∫ cos2 tdt = 1 2 ∫ (1 + cos 2t)dt = 1 2 ( t+ sin 2t 2 ) + C = 1 2 ( sin−1 ex + ex √ 1− e2x ) + C. 20. ∫ cos θ√ 2−sin2 θ dθ. Do the substitution sin θ = √ 2 sin t⇒ dθ = √ 2 cos tdt cos θ to get ∫ cos θ√ 2− sin2 θ dθ = ∫ cos θ√ 2− 2 sin2 t √ 2 cos t cos θ dt = ∫ dt = t+ C = sin−1 sin θ√ 2 + C. 21. ∫ 1 0 5x3 √ 1− x2dx. Do the substitution 1− x2 = t⇒ dx = − dt 2x 47 to get∫ 5x3 √ 1− x2dx = − ∫ 5x3 √ t dt 2x = −5 2 ∫ (1− t)√tdt = −5 2 ( 2 3 t3/2 − 2 5 t5/2 ) + C = −5 2 ( 2 3 √ (1− x2)3 − 2 5 √ (1− x2)5 ) + C. Hence∫ 1 0 5x3 √ 1− x2dx = −5 2 ( 2 3 √ (1− x2)3 − 2 5 √ (1− x2)5 )]1 0 = −2 3 . 22. ∫ 1/2 0 dx (1−x2)2 . Let x = sin t⇒ dx = cos tdt, hence ∫ dx (1− x2)2 = ∫ cos tdt cos4 t = ∫ sec3 tdt = 1 2 [sec t tan t− ln | sec t+ tan t|] + C. Therefore ∫ 1/2 0 dx (1− x2)2 = 1 2 [ x 1− x2 − ln 1 + x√ 1− x2 ]1/2 0 = 1 3 + 1 4 ln 3. 23. ∫ 2√ 2 dx x2 √ x2−1 . Let x = sec t⇒ dx = sec t tan tdt. Thus ∫ dx x2 √ x2 − 1 = ∫ sec t tan tdt sec2 t tan t = ∫ cos tdt = sin t+ C = √ 1− 1/x2 + C. 48 By plugging the limits √ 2 and 2 we get∫ 2 √ 2 dx x2 √ x2 − 1 = 1 2 ( √ 3− √ 2). 24. ∫ 2√ 2 √ 2x2−4 x dx. Let x = √ 2 sec t⇒ dx = √2 sec t tan tdt. Therefore∫ √ 2x2 − 4 x dx = ∫ 2 tan t√ 2 sec t √ 2 sec t tan tdt = 2 ∫ tan2 tdt = 2 ∫ (sec2 t− 1)dt = 2(tan t− t) + C = 2 (√ x2/2− 1− sec−1(x/ √ 2) ) + C. Thus ∫ 2 √ 2 √ 2x2 − 4 x dx = 2 [√ x2/2− 1− sec−1(x/ √ 2) ]2 √ 2 = 2(1− pi/4). 33. ∫ dx x2−4x+5dx. We complete the square for x 2 − 4x + 5 in order to use trigonometric substitution. x2 − 4x+ 5 = x2 − 4x+ 4− 4 + 5 = (x− 2)2 + 1. Therefore∫ dx x2 − 4x+ 5 = ∫ dx (x− 2)2 + 1 (let x− 2 = tan t) = ∫ sec2 tdt sec2 t = t+ C = arctan(x− 2) + C. 34. ∫ dx√ 2x−x2 . We complete the square for 2x− x2 to get∫ dx√ 2x− x2 = ∫ dx√1− (x− 1)2 (let x− 1 = sin t) = ∫ cos tdt cos t dt = t+ C = arcsin(x− 1) + C. 49 35. ∫ dx√ 3+2x−x2 .∫ dx√ 3 + 2x− x2 = ∫ dx√ 4− (x− 1)2 (let x− 1 = 2 sin t) = ∫ 2 cos tdt√ 4 cos2 t = t+ C = arcsin ( x− 1 2 ) + C. 36. ∫ dx 16x2+16x+5 .∫ dx 16x2 + 16x+ 5 = ∫ dx 16(x+ 1/2)2 + 1 (let x+ 1/2 = tan t/4) = 1 4 ∫ sec2 tdt sec2 t = 1 4 t+ C = 1 4 arctan [4(x+ 1/2)] + C. 1.5 Partial fractions For questions 1-8, write out the form of the partial fraction decomposition. 1. 3x− 1 (x− 3)(x+ 4) = A x− 3 + B x+ 4 . 2. 5 x(x2 − 4) = 5 x(x− 2)(x+ 2) = A x + B x− 2 + C x+ 2 . 3. 2x− 3 x3 − x2 = 2x− 3 x2(x− 1) = A x + B x2 + C x− 1 . 50 4. x2 (x+ 2)3 = A x+ 2 + B (x+ 2)2 + C (x+ 2)2 . 5. 2x− 3 x3(x2 + 2) = A x + B x2 + C x3 + Dx+ E x2 + 2 . 6. 3x (x− 1)(x2 + 6) = A x− 1 + Bx+ C x2 + 6 . 7. 4x3 − x (x2 + 5)2 = Ax+B (x2 + 5) + Cx+D (x2 + 5)2 . 8. 1− 3x4 (x− 2)(x2 + 1)2 = A x− 2 + Bx+ C x2 + 1 + Dx+ E (x2 + 1)2 . 51 Evaluate the following integrals: 9. ∫ dx x2−3x−4 . We begin by writing the partial fraction decomposition for 1/(x2 − 3x− 4). 1 x2 − 3x− 4 = 1 (x− 4)(x+ 1) = A x− 4 + B x+ 1 ⇒ 1 = A(x+ 1) +B(x− 4). In order to find A and B we plug two values for x: x = 4⇒ 1 = 5A⇒ A = 1 5 , x = −1⇒ 1 = −5B ⇒ B = −1 5 . Therefore ∫ dx x2 − 3x− 4 = ∫ ( 1/5 x− 4 − 1/5 x+ 1 ) dx = 1 5 ln |x− 4| − 1 5 ln |x+ 1|+ C = 1 5 ln ∣∣∣∣x− 4x+ 1 ∣∣∣∣+ C. 10. ∫ dx x2−6x−7 . We do as above: 1 x2 − 6x− 7 = 1 (x− 7)(x+ 1) = A x− 7 + B x+ 1 ⇒ 1 = A(x+ 1) +B(x− 7). x = −1⇒ 1 = −8B ⇒ B = −1 8 ;x = 7⇒ 1 = 8A⇒ A = 1 8 . Hence ∫ dx x2 − 6x− 7 = ∫ ( 1/8 x− 7 − 1/8 x+ 1 ) dx = 1 8 ln |x− 7| − 1 8 ln |x+ 1|+ C = 1 8 ln ∣∣∣∣x− 7x+ 1 ∣∣∣∣+ C. 11. ∫ 11x+17 2x2+7x−4dx. Applying same ideas as above, we get∫ 11x+ 17 2x2 + 7x− 4dx = ∫ ( 3 x+ 4 + 5 2x− 1 ) dx = 3 ln |x+ 4|+ 5 2 ln |2x− 1|+ C. 52 12. ∫ 5x−5 3x2−8x−3dx.∫ 5x− 5 3x2 − 8x− 3dx = ∫ ( 1 x− 3 + 2 3x+ 1 ) dx = ln |x− 3|+ 2 3 ln |3x+ 1|+ C. 13. ∫ 2x2−9x−9 x3−9x dx. We observe that 2x2 − 9x− 9 x3 − 9x = 2x2 − 9x− 9 x(x− 3)(x+ 3) = A x + B x− 3 + C x+ 3 which means 2x2 − 9x− 9 = A(x− 3)(x+ 3) +Bx(x+ 3) + Cx(x− 3). To find the values of A,B and C we plug x = 0⇒ −9 = −9A⇒ A = 1. x = 3⇒ −18 = 18B ⇒ B = −1. x = −3⇒ 36 = 18C ⇒ C = 2. Therefore∫ 2x2 − 9x− 9 x3 − 9x dx = ∫ ( 1 x − 1 x− 3 + 2 x+ 3 ) dx = ln |x| − ln |x− 3|+ 2 ln |x+ 3|+ C. 14. ∫ dx x(x2−1) . 1 x(x2 − 1) = A x + B x− 1 + C x+ 1 ⇒ 1 = A(x2 − 1) +Bx(x+ 1) + Cx(x− 1) = A = −1(x = 0);B = 1 2 (x = 1);C = 1 2 (x = −1). ∫ dx x(x2 − 1) = ∫ (−1 x + 1 2(x− 1) + 1 2(x+ 1) ) dx = − ln |x|+ 1 2 ln |x− 1|+ 1 2 ln |x+ 1|+ C. 53 15. ∫ x2−8 x+3 dx. Since the degree of the numerator is larger than the degree of the denominator, we use long division to get∫ x2 − 8 x+ 3 = ∫ ( x− 3 + 1 x+ 3 ) dx = x2 2 − 3x+ ln |x+ 3|+ C. 16. ∫ x2+1 x−1 dx. Use long division as above to get∫ x2 + 1 x− 1 dx = ∫ ( x+ 1 + 2 x− 1 ) dx = x2 2 + x+ 2 ln |x− 1|+ C. 17. ∫ 3x2−10 x2−4x+4dx. Use long division:∫ 3x2 − 10 x2 − 4x+ 4dx = ∫ ( 3 + 12x− 22 x2 − 4x+ 4 ) dx. Now use partial fractions to simplify: 12x− 22 x2 − 4x+ 4 = 12x− 22 (x− 2)2 = A x− 2 + B (x− 2)2 ⇒ 12x− 22 = A(x− 2) +B ⇒ B = 2(x = 2)⇒ −10 = −A+B(x = 1)⇒ A = 12. Hence ∫ 3x2 − 10 x2 − 4x+ 4dx = ∫ ( 3 + 12 x− 2 + 2 (x− 2)2 ) dx = 3x+ 12 ln |x− 2| − 2 x− 2 + C. 18. ∫ x2 x2−3x+2dx. Since the degree of the top is the same as the degree of the bottom we use long division: x2 x2 − 3x+ 2 = 1 + 3x− 2 x2 − 3x+ 2 = 1 + A x− 2 + B x− 1 . 54 We find A and B: 3x− 2 x2 − 3x+ 2 = A x− 2 + B x− 1 ⇒ 3x− 2 = A(x− 1) +B(x− 2) ⇒ B = −1(x = 1), A = 4(x = 2). Therefore∫ x2 x2 − 3x+ 2dx = ∫ ( 1 + 4 x− 2 − 1 x− 1 ) dx = x+ 4 ln |x− 2| − ln |x− 1|+ C. 19. ∫ x5+x2+2 x3−x dx. Long division gives: x5 + x2 + 2 x3 − x = x 2 + 1 + x2 + x+ 2 x3 − x and x2 + x+ 2 x3 − x = A x + B x− 1 + C x+ 1 ⇒ x2 + x+ 2 = A(x2 − 1) +Bx(x+ 1) + Cx(x− 1) ⇒ (x = 0⇒ A = −2), (x = 1⇒ B = 2), (x = −1⇒ C = 1). Hence∫ x5 + x2 + 2 x3 − x dx = ∫ ( x2 + 1 + −2 x + 2 x− 1 + 1 x+ 1 ) dx = x3 3 + x− 2 ln |x|+ 2 ln |x− 1|+ ln |x+ 1|+ C. 20. ∫ x5−4x3+1 x3−4x dx. Imitate the above solution to get∫ x5 − 4x3 + 1 x3 − 4x dx = ∫ ( x2 + 1 x3 − 4x ) dx = ∫ ( x2 + A x + B x− 2 + C x+ 2 ) dx = x3 3 + A ln |x|+B ln |x− 2|+ C ln |x+ 2|+D, 55 where A,B and C may be found by: 1 x3 − 4x = A x + B x− 2 + C x+ 2 ⇒ 1 = A(x2 − 4) +Bx(x+ 2) + Cx(x− 2) ⇒ (x = 0⇒ A = −1 4 ), (x = 2⇒ B = 1 8 ), (x = −2⇒ C = −1 8 ). 21. ∫ 2x2+3 x(x−1)2dx.∫ 2x2 + 3 x(x− 1)2dx = ∫ ( A x + B x− 1 + C (x− 1)2 ) dx = A ln |x|+B ln |x− 1| − C x− 1 +D, where A,B and C may be found by 2x2 + 3 = A(x− 1)2 +Bx(x− 1) + Cx ⇒ (x = 0⇒ A = −3), (x = 1⇒ C = 5) (x = −1⇒ 5 = −2A+ 2B − C ⇒ B = 2). 22. ∫ 3x2−x+1 x3−x2 dx.∫ 3x2 − x+ 1 x3 − x2 dx = ∫ ( A x + B x2 + C x− 1 ) dx = A ln |x| − B x + C ln |x− 1|+D, and we may find the constants A,B and C as following: 3x2 − x+ 1 = Ax(x− 1) +B(x− 1) + Cx2 ⇒ (x = 0⇒ B = −1), (x = 1⇒ C = 3), (x = −1⇒ 5 = 2A− 2B + C ⇒ A = 0). 23. ∫ 2x2−10x+4 (x+1)(x−3)2dx.∫ 2x2 − 10x+ 4 (x+ 1)(x− 3)2dx = ∫ ( A x+ 1 + B x− 3 + C (x− 3)2 ) dx = A ln |x+ 1|+B ln |x− 3| − C x− 3 +D, 56 where A,B and C may be found by: 2x2 − 10x+ 4 = A(x− 3)2 +B(x+ 1)(x− 3) + C(x+ 1) ⇒ (x = 3⇒ C = −2); (x = −1⇒ A = 1); (x = 0⇒ B = 1). 24. ∫ 2x2−2x−1 x3−x2 dx.∫ 2x2 − 2x− 1 x3 − x2 dx = ∫ ( A x + B x2 + C x− 1 ) dx = A ln |x| − B x + C ln |x− 1|+D, where A,B and C may be found by: 2x2 − 2x− 1 = Ax(x− 1) +B(x− 1) + Cx2 ⇒ (x = 0⇒ B = 1); (x = 1⇒ C = −1); (x = 2⇒ A = 3). 25. ∫ x2 (x+1)3 dx.∫ x2 (x+ 1)3 dx = ∫ ( A x+ 1 + B (x+ 1)2 + C (x+ 1)3 ) dx = A ln |x+ 1| − B x+ 1 − C 2(x+ 1)2 +D. A,B and C may be found as usual: x2 = A(x+ 1)2 +B(x+ 1) + C ⇒ (x = −1⇒ C = 1); (x = 0⇒ A+B + C = 0); (x = 1⇒ 1 = 4A+ 2B + C). Solving the above equations gives A = 1, B = −2, C = 1. 57 1.6 Improper integrals 1. In each part, determine whether the integral is improper, and if so, ex- plain why. (a) ∫ 5 1 dx x−3 . The integral is improper since 1/(x− 3), the integrand, has an infinite discontinuity at x = 3 which is a point of [1, 5]. (b) ∫ 5 1 dx x+3 is not an improper integral. The integrand is continuous on the interval of integration. (c) ∫ 1 0 lnxdx is an improper integral because lnx has an infinite dis- continuity at 0, namely, limx→0+ lnx = −∞. (d) ∫∞ 1 e−xdx is an improper integral because ∞ is a limit of the inte- gral. (e) ∫∞ −∞ dx 3√x−1 is an improper integral because ∞ and −∞ are limits of the integral. (f) ∫ pi/4 0 tan xdx is not an improper integral because tanx is continuous on [0, pi/4]. 2. In each part, determine the value of p for which the integral is improper: (a) ∫ 1 0 1 xp dx. We do the integral: For p < 1,∫ 1 0 1 xp dx= lim b→0+ ∫ 1 b x−pdx = lim b→0+ x1−p 1− p ]1 b = lim b→0+ ( 1 1− p − b1−p 1− p ) = 1 1− p. Thus, if p < 1 the integral is not improper. 58 For p > 1 the above limit is infinite, thus the integral is improper. Also, if p = 1 the integral will be improper because limb→0+ ln b = −∞. Consequently, the values of p which make the above integral im- proper are {p ≥ 1}. (b) ∫ 2 1 dx x−p . This integral will be improper if x−p = 0 for some p ∈ [1, 2]. Hence, p ∈ [1, 2]. (c) ∫ 1 0 e−pxdx. Since e−px is continuous on [0, 1] for all p, this integral is always proper (not improper.) Evaluate the integrals that converge: 3. ∫∞ 0 e−2xdx, we evaluate:∫ e−2xdx = −1 2 e−2x + C, hence lim b→∞ ∫ b 0 e−2xdx = lim b→∞ −1 2 e−2x ]b 0 = lim b→∞ ( 1 2 − 1 2 e−2b ) = 1 2 . Therefore, the integral converges and is equal to 1/2. 4. ∫∞ −1 dx 1+x2 . We have lim b→∞ ∫ b −1 dx x2 + 1 = lim b→∞ tan−1 x ]b −1 = lim b→∞ ( tan−1 b− tan−1(−1)) = pi 2 − −pi 4 = 3pi 4 . 5. ∫∞ 3 2 x2−1dx. Do the integral by partial fractions to get∫ 2 x2 − 1dx = ln ∣∣∣∣x− 1x+ 1 ∣∣∣∣+ C. 59 Hence, lim b→∞ ∫ b 3 2 x2 − 1dx = limb→∞ ln ∣∣∣∣x− 1x+ 1 ∣∣∣∣]b 3 = lim b→∞ ln ∣∣∣∣b− 1b+ 1 ∣∣∣∣− ln 12 = ln 1 + ln 2 = ln 2. Observe that L’Hopital’s rule has been used to evaluate the above limit. 6. ∫∞ 0 xe−x 2 dx. Do the indefinite integral by substitution to get:∫ xe−x 2 dx = −1 2 e−x 2 + C. Hence lim b→∞ ∫ b 0 xe−x 2 dx = lim b→∞ −1 2 e−x 2 ]b 0 = lim b→∞ ( 1 2 − 1 2 e−b 2 ) = 1 2 . 7. ∫∞ e 1 x ln3 x dx. Do the indefinite integral by substitution: ln x = t to get∫ 1 x ln3 x dx = −1 2 ln2 x + C. Hence, lim b→∞ ∫ b e 1 x ln3 x dx = lim b→∞ −1 2 ln2 x ]b e = lim b→∞ ( 1 2 − 1 2 ln2 b ) = 1 2 . 8. ∫∞ 2 dx x √ lnx . Let lnx = t to get∫ dx x √ lnx = 2 √ lnx+ C. Hence lim b→∞ ∫ b 2 1 x √ lnx dx = lim b→∞ 2 √ lnx ]b 2 (1.6.1) = lim b→∞ ( 2 √ ln b− 2 √ ln 2 ) =∞. (1.6.2) Thus the integral diverges to ∞. 60 9. ∫ 0 −∞ dx (2x−1)3 . lim b→−∞ ∫ 0 b dx (2x− 1)3 = limb→−∞ − 1 4(2x− 1)2 ]0 b = lim b→−∞ ( 1 4(2b− 1)2 − 1 4 ) = −1 4 . 10. ∫ 3 −∞ dx x2+9 . lim b→−∞ ∫ 3 b dx x2 + 9 = lim b→−∞ 1 3 tan−1 ( 1 3 x )]3 b = 1 3 lim b→−∞ ( tan−1 1− tan−1 b 3 ) = 1 3 ( pi 4 − −pi 2 ) = pi 4 . 11. ∫ 0 −∞ e 3xdx. lim b→−∞ ∫ 0 b e3xdx = lim b→−∞ e3x 3 ]0 b = lim b→−∞ ( 1 3 − e 3b 3 ) = 1 3 . 12. ∫ 0 −∞ ex 3−2exdx. Observe first that the only problem is −∞. That is, 3 − 2ex = 0 gives a solution that does not lie in (−∞, 0]. lim b→−∞ ∫ 0 b ex 3− 2exdx 61
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