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#