How do you find the integral of $int sin^5(x)cos^8(x) dx$ ?

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Answer 1 · 2015-09-27T05:10:29

$-cos^9x/9 + (2cos^11x)/11 - cos^13x/13 + C$

$I=int sin^5x cos^8xdx=intsinx sin^4x cos^8x dx$

$I=int sinx (sin^2x)^2 cos^8xdx=int (1-cos^2x)^2 cos^8x sinx dx$

$cosx=t => -sinxdx=dt => sinxdx=-dt$

$I=int (1-t^2)^2 t^8 (-dt) = -int (1-2t^2+t^4)t^8 dt$

$I=-int (t^8-2t^10+t^12) dt = -t^9/9+(2t^11)/11-t^13/13+C$

$I=-cos^9x/9 + (2cos^11x)/11 - cos^13x/13 + C$

How do you find the integral of int sin^5(x)cos^8(x) dxint sin^5(x)cos^8(x) dx?