How do you find the integral of $int [cot^5 (x) (sin^4(x) dx]$ ?

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Answer 1 · 2015-10-07T15:08:41

$I=ln|sinx|-sin^2x+sin^4x/4+C$

$int cot^5xsin^4xdx=int (cos^5x)/(sin^5x)sin^4xdx=$

$int (cos^4xcosx)/(sinx)dx=int ((1-sin^2x)^2cosx)/sinx dx=I$

$sinx=t => cosxdx=dt$

$I=int (1-t^2)^2/tdt=int (1-2t^2+t^4)/tdt=int (1/t-2t+t^3)dt$

$I=ln|t|-t^2+t^4/4+C$

$I=ln|sinx|-sin^2x+sin^4x/4+C$

How do you find the integral of int [cot^5 (x) (sin^4(x) dx]int [cot^5 (x) (sin^4(x) dx]?