How do you find the integral of int cos^5xsin^4xdx∫cos5xsin4xdx?

1 Answer
Konstantinos Michailidis
Sep 18, 2015

Refer to explanation

Explanation:

We have that

int cos^5xsin^4xdx=intcosx*cos^4x*sin^4xdx= intcosx(1-sin^2x)*(1-sin^2x)*sin^4xdx∫cos5xsin4xdx=∫cosx⋅cos4x⋅sin4xdx=∫cosx(1−sin2x)⋅(1−sin2x)⋅sin4xdx

We set sinx=tsinx=t hence dxcosx=dtdxcosx=dt so we have

int(1-t^2)(1-t^2)*t^4dt=int(t^4+t^8-2t^6)dt= t^5/5+t^9/9-2y^7/7+c∫(1−t2)(1−t2)⋅t4dt=∫(t4+t8−2t6)dt=t55+t99−2y77+c

Hence we replace with sinxsinx to get

int cos^5xsin^4xdx=sin^5x/5+sin^9x/9-2*sin^7x/7+c∫cos5xsin4xdx=sin5x5+sin9x9−2⋅sin7x7+c

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