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How do you integrate #sin^2xcosx dx#?

1 Answer

Feb 4, 2015

I would use the fact that:

#d(sin(x))=cos(x)dx# and substituting:

#intsin^2(x)*d(sin(x))=#

Now I use #sin(x)# as my variable of integration and integrate it (as if it was a simple #x#).

#intsin^2(x)*d(sin(x))=sin^3(x)/3+c#

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