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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Help simplifying/checking integral

**Physics Forums | Science Articles, Homework Help, Discussion**