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,

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

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

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

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

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

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