sin^3xcos^2x
after messing around a bit I did this
= sinx(sin^2x)(cos^x)
= sinx(1-cos^2x)(cos^2)
= sinx(cos^2x - cos^4x)
= sinxcos^2x - sinxcos^4x
and the integral of that is
(-1/3)cos^3x + (1/5)cos^5x + C
the integral of sin^3xcos^2x
my answer is -(1/2)(cosx)^2+(1/4)(cosx)^4+c but this is incorrect, please help.
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