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7.4. Trigonometric Integrals www.ck12.org 7.4 Trigonometric Integrals 1. Let u = cos x. Then du = − sin x dxx. Z cos4 x sin x dx = − Z u4 du u5 +C 5 cos5 x +C =− 5 =− 2. Let u = 5ϕ. Then du = 5dϕ. Z sin2 5ϕdϕ = = = = = 1 sin2 u du 5 Z 1 1 (1 − cos 2u)du 5 2 1 1 u − sin 2u +C 10 2 1 1 5ϕ − sin(2 × 5ϕ) +C 10 2 1 1 ϕ − sin(10ϕ) +C 2 20 Z 3. Let u = sin 2z. Then du = 2 cos 2z dz. Z sin2 2z cos3 2z dz = Z sin2 2z cos3 2z(1 − sin2 2z)dz Z 2 sin 2z cos 2z dz − = Z sin4 2z cos 2z dz 1 1 u2 du − u4 du 2 2 u3 u5 = − +C 6 10 sin3 2z sin5 2z = − +C 6 10 Z Z = 4. Z sin x cos Let u = cos 234 x 2 . Then du = − 12 sin x 2 x dx. 2 x x cos cos dx 2 2 2 Z x x = 2 cos2 sin dx 2 2 Z dx = 2 sin x www.ck12.org Chapter 7. Integration Techniques, Solution Key Z 2 cos 2 x 2 sin x 2 dx = −4 Z u2 du 4u3 +C 3 x 4 = − cos3 +C 3 2 =− 5. Let u = sec x. Then du = sec x tan x dx. Z sec4 x tan3 xdx = Z tan2 x sec3 x(sec x tan x)dx Z (sec2 x − 1) sec3 x(sec x tan x)dx Z sec5 x(sec x tan x)dx − Z u5 du − = = = Z Z sec3 x(sec x tan x)dx u3 du u6 u4 − +C 6 4 sec6 x sec4 x = − +C 6 4 = 6. Z tan4 x sec xdx = = Z (sec2 x − 1)(sec2 x − 1) sec x dx Z (sec5 x − 2 sec3 x + sec x)dx sec3 x tan x 3 sec x tan x 1 + sec3 x dx − 2 −2× sec x dx + ln|sec x + tan x|+C 4 4 2 2 Z sec x tan x sec3 x tan x 3 sec x tan x 1 + + sec x dx − 2 − ln|sec x + tan x|+ − ln|sec x + tan x|+C = 4 4 2 2 2 sec3 x tan x 5 sec x tan x 3 = − + ln|sec x + tan x| 4 8 8 Z Z = 7. Z √ tan x sec4 x dx = Z Z 1 tan 2 x sec4 x dx 1 tan 2 x(tan2 x + 1) sec2 xdx Z h i 5 1 = tan 2 x sec2 x + tan 2 x sec2 x dx = Let u = tan x. Then du = sec2 x dx. 235 7.4. Trigonometric Integrals www.ck12.org Z h Z Z i 5 1 5 1 tan 2 x sec2 x + tan 2 x sec2 x dx = u 2 du + u 2 du 7 = u2 7 2 3 + u2 3 2 7 +C 3 2 tan 2 x 2 tan 2 x + +C = 7 3 8. Let u = 2x . Then du = 12 dx. If x = 0, then u = 0. If x = π2 , then u = π4 . π Z2 π tan5 x 2 Z4 dx = 0 2 tan5 u du 0 π = 2 tan4 u π 4 4 −2 Z4 0 tan3 u du 0 π √ !4 Z2 2 2 tan4 u π4 − =2 + 2 tan u du 2 2 0 0 √ !2 π 2 8 − = − 2ln|cos u|04 16 2 π 8 = − 1 − 2ln cos −0 16 4 √ 2 1 = − − 2ln 2 2 4 1 = − + ln 2 2 1 = − + ln2 2 9. 236 www.ck12.org Chapter 7. Integration Techniques, Solution Key π π Z4 Z4 V = π cos2 x dx − 0 sin2 x dx 0 π π 1 x 4 x 4 1 cos x sin x + − π − cos x sin x + 2 2 2 2 0 0 1 π 1 π + − − + =π 4 8 4 8 1 =π 2 π = 2 =π 10. a. R x+cot x csc x dx = csc x csc csc x+cot x dx R Let u = csc x + cot x. Then du = (− csc x cot x − csc2 x)dx. Z 1 du u = −ln|u|+C csc xdx = − Z = −ln|csc x + cot x|+C b. Z 1 dx sin x Z sin x = dx 2 sin x Z Z csc x dx = = sin x dx (1 − cos x)(1 + cos x) Let u = cos x. Then du = − sin x dx. 237 7.4. Trigonometric Integrals Z www.ck12.org sin x dx = − (1 − cos x)(1 + cos x) du (1 − u)(1 + u) Z Z A B = du + du 1−u 1+u =− Z = Z Z 1 2 1−u du − 1 2 1 − cos x Z du − 1 2 1+u Z du 1 2 1 + cos x du 1 1 = |1 − cos x|− |1 + cos x|+C 2 2 1 1 − cos x = +C 2 1 + cos x r 1 − cos x +C = ln x 1+xcos = ln tan +C 2 OR r ln 1 − cos x +C = ln 1 + cos x s s (1 − cos x) (1 − cos x) · +C (1 + cos x) (1 − cos x) (1 − cos x)2 +C 1 − cos2 x 1 − cos x +C = ln sin x 1 cos x = ln − +C sin x sin x = ln|csc x − cot x|+C = ln 238