How do I evaluate the indefinite integral intsin^3(x)*cos^2(x)dxsin3(x)cos2(x)dx ?

1 Answer
Sep 21, 2014

The answer is -(cos^3x)/3+(cos^5x)/5+Ccos3x3+cos5x5+C.

The trick with sinusoidal powers is to use identities so that you can have sin xsinx or cos xcosx with a power of 1 and use substitution.

In this case, it is easier to get sin xsinx to a power of 1 using sin^2x=1-cos^2xsin2x=1cos2x.

int sin^3x*cos^2x dxsin3xcos2xdx
=int sin x(1-cos^2x)cos^2x dx=sinx(1cos2x)cos2xdx
=int sin x(cos^2x-cos^4x)dx=sinx(cos2xcos4x)dx
=int sin x cos^2xdx-int sin x cos^4x dx=sinxcos2xdxsinxcos4xdx

Now it is a matter of using substitution:

u=cos xu=cosx
du = -sin x dxdu=sinxdx

int sin x cos^2xdx-int sin x cos^4x dxsinxcos2xdxsinxcos4xdx
=int -u^2 du+ int u^4 du=u2du+u4du
=-(u^3)/3+(u^5)/5+C=u33+u55+C
=-(cos^3x)/3+(cos^5x)/5+C=cos3x3+cos5x5+C