OK, so I'm just trying to refresh my calc knowledge before I go off to school. The problem I'm trying to do is:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\int sin(2x)tan^{-1}(sin(x))dx[/itex]

So what I've done is, first, replace sin(2x), giving:

[itex]\int 2cos(x)sin(x)tan^{-1}(sin(x))dx[/itex]

or

[itex]2\int cos(x)sin(x)tan^{-1}(sin(x))dx[/itex]

Then, I assume that sin(x) = u (and therefore, du = cos(x)dx). This gives me:

[itex]2\int utan^{-1}(u)du[/itex]

Then, I integrate by parts.

[itex]2\int utan^{-1}(u)du = 2(\frac{1}{2}u^{2}tan^{-1}(u)-\int \frac{u^{2}}{2+2u^{2}})[/itex]

or

[itex]2(\frac{1}{2}u^{2}tan^{-1}(u)-(\frac{u^{3}}{6}+\frac{u}{2}))[/itex]

Substituting back in for u gives me:

[itex]2(\frac{1}{2}sin^{2}(x)tan^{-1}(sin(x))-(\frac{sin^{3}(x)}{6}+\frac{sin(x)}{2}))[/itex]

Or, to simplify a little:

[itex]-\frac{sin^{3}(x)}{3}+tan^{-1}(sin(x))sin^{2}(x)-sin(x)[/itex]

My question is where do I go from here to simplify, and also if the work I've done up to this point is correct/on the right track (it's been a while since I took calc so go easy hahah). Just so you all know, I don't have an answer key for this problem so I'm working off of Wolfram Alpha's answer which is:

[itex]\frac{1}{2}((cos(2x)-3)(-tan^{-1}(sin(x)))-2sin(x)) + C [/itex]

Thanks in advance!

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# Homework Help: Help simplifying/checking integral

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