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The problem and my attempt to solve is as follows.

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*cos^3(x) dx\)

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*cos^2(x)*cos(x) dx\)

Using a Pythagorean identity,

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*(1-sin^2(x))*cos(x) dx\)

Distributing

\(\displaystyle \int[\sin^{\frac{-3}{2}}(x)-sin^{-3}(x)]*cos(x)dx\)

Substituting

U=sin(x) du=cos(x)dx

\(\displaystyle \int [u^{\frac{-3}{2}}-u^{-3}]du\)

Integrating

\(\displaystyle -2u^{\frac{-1}{2}}+\frac{1}{2}u^{-2}\)

And finally, and simplified as I can see,

\(\displaystyle -\frac{2}{\sqrt{\sin(x)}}+\frac{1}{2\sin^2(x)}+C\)

I've looked at this a few times and couldn't find any errors, but Wolfram is coming up with an answer that doesn't Jive with what I have. That being the case, I would really appreciate it if someone could take a look at this problem and let me know where I went wrong.

Thanks much,

Mac