Trigonometric Integrals 1. β« π ππ3 π₯ πππ 2 π₯ ππ₯ = β« sin π₯ π ππ2 π₯ πππ 2 π₯ ππ₯ split the odd trig power sin3x into sin x sin2x = β« sin π₯ (1 β πππ 2 π₯) πππ 2 π₯ ππ₯ replace sin2x = 1 β cos2 x 2 2 = β« sin π₯ (1 β πππ π₯) πππ π₯ ππ₯ the integral now has a single sin x 2 2 = β«(1 β π’ ) π’ ππ’ Use the substitution u = cos x 2 4 = β« π’ β π’ ππ’ du = sin x dx 1 3 1 5 = 3π’ β 5π’ + πΆ 1 = 3 π 1 π ππ3 π₯ β 5 π ππ5 π₯ + πΆ replace the u with sin x π 2. β«02 π ππ5 π₯ πππ 7 π₯ ππ₯ = β«02 sin π₯ π ππ4 π₯ πππ 7 π₯ ππ₯ split the odd trig power sin5x into sin x sin4x π 2 = β«0 sin π₯ (1 β πππ 2 π₯)2 πππ 7 π₯ ππ₯ replace sin4x = (sin2x)2 = (1 β cos2 x)2 = β«02 sin π₯ (1 β πππ 2 π₯)2 πππ 7 π₯ ππ₯ the integral now has a single sin x π 0 = β«1 (1 β π’2 )2 π’7 ππ’ Use the substitution u = cos x du = sin x dx when x = 0 u = cos (0) = 1 0 = β«1 (1 β 2π’2 + π’4 ) π’7 ππ’ 1 = β«0 π’7 β 2π’9 + π’11 ππ’ when x = π2 u = cos(π2) = 0 π’=0 π’=1 1 1 1 = (0) β (8 (1)8 β 5 (1)10 + 12 (1)12 ) 1 = 8 1 1 π’8 β 5 π’10 + 12 π’12 ] 1 = β 120 1 3. β« π ππ2 (ππ₯) πππ 5 (ππ₯) ππ₯ = β« π π ππ2 π€ πππ 5 π€ ππ€ Use the substitution w 1 = β« π π ππ2 π€ πππ 4 π€ cos π₯ ππ€ 1 = π 1 = π 2 dw 4 = ππ₯ = πππ₯ 1 π β« π ππ π€ πππ π€ cos π₯ ππ€ dw = ππ₯ β« π ππ2 π€ (1 β π ππ2 π€)2 cos π₯ ππ€ We can now use the substitution u = sin w π 4. β«02 π ππ2 π₯ ππ₯ π 1 = β«02 2 (1 β πππ 2π₯) ππ₯ replace sin2x = ½ (1 β cos (2x)) = β«02 (2 β 2 πππ 2π₯) ππ₯ You can now integrate π 1 1 π 5. β«02 π ππ4 π₯ ππ₯ π β«02 [π ππ2 π₯ ]2 ππ₯ replace sin4x = (sin2x)2 = β«0 [2 (1 β πππ 2π₯)] 2 ππ₯ replace sin2x = ½ (1 β cos (2x)) = β«02 4 [(1 β πππ 2π₯)] 2 ππ₯ = β«02 4 (1 β 2πππ 2π₯ + πππ 2 2π₯)ππ₯ = β«02 4 β 2 πππ 2π₯ + 4 πππ 2 2π₯ ππ₯ = β«02 4 β 2 πππ 2π₯ + 4 2 (1 + πππ 4π₯) ππ₯ replace cos22x = ½ (1 + cos (4x)) = β«02 4 β 2 πππ 2π₯ + 8 + 8 πππ 4π₯ ππ₯ = β«02 8 β 2 πππ 2π₯ + 8 πππ 4π₯ ππ₯ = π 2 1 π 1 π 1 π 1 1 1 π 1 1 11 π 1 1 1 π 3 1 1 π 1 You can now integrate π 6. β«02 π ππ2 π₯ πππ 2 π₯ ππ₯ = β«02 (π πππ₯ πππ π₯)2 ππ₯ π 2 1 = β«02 (2 π ππ2π₯) ππ₯ π 2 1 π ππ2 2π₯ ππ₯ 4 π 11 β«02 4 2 (1 β πππ 4π₯) π 1 1 β«02 8 β 8 πππ 4π₯ ππ₯ = β«0 = = replace sin x cos x = ½ sin (2x) ππ₯ 1 7. β« π₯πππ 2 π₯ ππ₯ = β« π₯ (1 + πππ 2π₯) ππ₯ 2 1 replace sin22x = ½ (1 - cos (4x)) You can now integrate replace cos2x = ½ (1 + cos (2x)) 1 = β«(2 π₯ + 2 π₯ πππ 2π₯) ππ₯ 1 1 = β« 2 π₯ ππ₯ + β«(2 π₯ πππ 2π₯) ππ₯ 1 1 You can now integrate using integration by parts on β«(2 π₯ πππ 2π₯) ππ₯ with f(x) = 2 π₯ π ππ5 π₯ 8. β« πππ π₯ ππ₯ = β« β π ππ4 π₯ sin π₯ ππ₯ βπππ π₯ 2 (π ππ2 π₯) sin π₯ = β« = β« = β« = β« split sin5x into sin4x sin x ππ₯ βπππ π₯ 2 (1βπππ 2 π₯) sin π₯ βπππ π₯ 2 (1βπ’2 ) ππ’ βπ’ 1β2π’2 +π’4 = β« (π’ 1 π’2 1 β 2 ππ₯ Use the substitution u = cos x ππ’ 3 ` 7 β 2π’2 + π’2 ) ππ’ You can now integrate β remember to replace the u = cos x du = sinx dx 9. β« πππ 2 π₯ π‘ππ5 π₯ ππ₯ π ππ5 π₯ = β« πππ 2 π₯ = β« = β« = β« = β« = β« ππ₯ πππ 5 π₯ π ππ4 π₯ π πππ₯ πππ 3 π₯ ππ₯ (π ππ2 π₯)2 π πππ₯ πππ 3 π₯ ππ₯ (1βπππ 2 π₯)2 π πππ₯ πππ 3 π₯ (1βπ’2 )2 π’3 ππ₯ ππ’ 1β2π’2 +π’4 π’3 ππ’ 2 = β« (π’β3 β π’ + π’) ππ’ 1 1 = β 2 π’β2 β 2 ln(π’) + 2 π’2 1 1 = β 2π’2 β 2 ln(π’) + 2 π’2 1 1 = β 2πππ 2 π₯ β 2 ln(πππ ) + 2 πππ 2 π₯ 1 1 = β 2 π ππ 2 π₯ β 2 ln(πππ ) + 2 πππ 2 π₯ + πΆ 10. β« tan π₯ π ππ 3 π₯ ππ₯ = β« tan π₯ π ππ 2 π₯ π πππ₯ ππ₯ = β« π’2 ππ’ 1 = 3 π’3 + C = 1 3 split the odd sec3x into sec x sec2x Use the substitution u = π πππ₯ du = sec x tan x π ππ 3 π₯ + C 11. β« π‘ππ2 π₯ ππ₯ = β«(π ππ 2 π₯ β 1) ππ₯ = tan x β x + C 12. β« sec π₯ π‘ππ3 π₯ ππ₯ = β« sec π₯ tan π₯ π‘ππ2 π₯ ππ₯ = β« sec π₯ tan π₯ (π ππ 2 π₯ β 1) ππ₯ = β«(π’2 β 1) ππ’ 1 = 3 π’3 β π’ = 1 3 π ππ 3 π₯ β π πππ₯ + πΆ replace tan2x = sec2x β 1 Use the substitution u = π πππ₯ du = sec x tan x
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