How do you find the integral of $int cos^3x sin^2x dx$ ?

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Answer 1 · 2015-09-30T07:51:36

$intcos^3xsin^2xdx=(1/3)sin^3x-(1/5)*sin^5x+c$

$intcos^3xsin^2xdx$

$=intcos^2x*sin^2x*(cosx)dx$

$=int (1-sin^2)sin^2x*d(sinx)$

Hints:

$(d(sinx))/dx=cosx$

$d(sinx)=cosxdx$

Therefore,

$=int(sin^2x-sin^4x)*d(sinx)$
$==(1/3)sin^3x-(1/5)*sin^5x+c$

How do you find the integral of int cos^3x sin^2x dxint cos^3x sin^2x dx?