How do I evaluate #int cos^5(x) sin^4(x) dx#?

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Answer 1 · 2018-04-20T09:56:28

#I=sin^5x/5-(2sin^7x)/7+sin^9x/9+c#

Here,

#I=intcos^5xsin^4xdx#

#=intsin^4x(cos^2x)^2cosxdx#

#=intsin^4x(1-sin^2x)^2cosxdx#

Let, #sinx=t=>cosxdx=dt#

So,

#I=intt^4(1-t^2)^2dt#

#=int(t^4(1-2t^2+t^4)dt#

#=int(t^4-2t^6+t^8)dt#

#=t^5/5-2(t^7/7)+t^9/9+c..towhere,t=sinx#

#=sin^5x/5-(2sin^7x)/7+sin^9x/9+c#