What is the integral of #int sin^5 (x) dx#?

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Answer 1 · 2018-03-04T14:56:16

The answer is #=-1/5cos^5x+2/3cos^3x-cosx+C#

We need

#sin^2x+cos^2x=1#

The integral is

#intsin^5dx=int(1-cos^2x)^2sinxdx#

Perform the substitution

#u=cosx#, #=>#, #du=-sinxdx#

Therefore,

#intsin^5dx=-int(1-u^2)^2du#

#=-int(1-2u^2+u^4)du#

#=-intu^4du+2intu^2du-intdu#

#=-u^5/5+2u^3/3-u#

#=-1/5cos^5x+2/3cos^3x-cosx+C#