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Integrate sin^5 (x) and similarly cos^5 (x).

Kindly refer the links for answer similar to your queries:

https://www.meritnation.com/ask-answer/question/integration-of-sin5xdx/integrals/6512709

(refer Q no. 11)
 https://www.meritnation.com/cbse-class-12-commerce/math/math-part-ii---ncert-solutions/integrals/ncert-solutions/59_1_1384_236_307_7510

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Hello Shanaya dear, ​∫ sin^5 x dx = ∫ sin^4 x * sin x dx
==> ∫(1 - cos^2 x)^2 * sin x dx
Plug cos x = t
So - sin x dx = dt
Plugging
==> ∫(1 - t^2)^2 * (-dt)
==> ∫ (1 + t^4 - 2t^2) * (-dt)
==> ∫ (2t^2 - t^4 - 1) dt
==> 2 t^3 /3 - t^5 /5 - t + C
Re plugging we have
2/3 * cos^3 x - 1/5 * cos^5 x - cos x + C
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sin2x + cos2 x = 1
Integral of sin5x.dx = Integral of (1-cos2x)(1-cos2x)(sinx).dx
= Integral of (1-cos2x)2(sinx).dx
let t = cosx
    dt= sinx.dx
Thus, 
Integral of (1-t2)2.dt
= Integral of (1 -2t​2 + t4).dt
= t - 2t3 +  t5 +c'
         3      5 

Sub the value of t as cosx in the above answer which will be your final answer.
Similarly for cos5x you need you convert it as integral of (1-sin2x)(1-sin2x)(cosx).dx and you will substituite t = sinx and proceed with the same way.
Hope its clear :)
 
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