Help simplifying/checking integral

[c.dube]

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:

$\int sin(2x)tan^{-1}(sin(x))dx$

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

$\int 2cos(x)sin(x)tan^{-1}(sin(x))dx$

or

$2\int cos(x)sin(x)tan^{-1}(sin(x))dx$

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

$2\int utan^{-1}(u)du$

Then, I integrate by parts.

$2\int utan^{-1}(u)du = 2(\frac{1}{2}u^{2}tan^{-1}(u)-\int \frac{u^{2}}{2+2u^{2}})$

or

$2(\frac{1}{2}u^{2}tan^{-1}(u)-(\frac{u^{3}}{6}+\frac{u}{2}))$

Substituting back in for u gives me:

$2(\frac{1}{2}sin^{2}(x)tan^{-1}(sin(x))-(\frac{sin^{3}(x)}{6}+\frac{sin(x)}{2}))$

Or, to simplify a little:

$-\frac{sin^{3}(x)}{3}+tan^{-1}(sin(x))sin^{2}(x)-sin(x)$

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:

$\frac{1}{2}((cos(2x)-3)(-tan^{-1}(sin(x)))-2sin(x)) + C$

Thanks in advance!

 

[Bohrok]

Then, I integrate by parts.

$2\int utan^{-1}(u)du = 2(\frac{1}{2}u^{2}tan^{-1}(u)-\int \frac{u^{2}}{2+2u^{2}})$

or

$2(\frac{1}{2}u^{2}tan^{-1}(u)-(\frac{u^{3}}{6}+\frac{u}{2}))$

There's an error from the integral on the right side of the top equation going to the line below it. Factor out a 2 from the denominator then try long division to make the integrand into something easier to integrate.

 

[c.dube]

Ahh I knew it was something really dumb like that! Thanks, but new question now. For the long division, what answer should I be getting? Because trying to work it through I'm not getting the right answer. I get an answer of $1-\frac{1}{u}$ and the remainder but I can't get the remainder (unless its $\frac{1}{u^{3}+u}$). The math doesn't seem to work when I check this can you help me out? Polynomial long division was always a weakness.

 

[c.dube]

Whoops never mind I got it ($1-\frac{1}{u^{2}+1}$) dumb mistake.

 

[Bohrok]

You can read about polynomial long division here
http://www.sosmath.com/algebra/factor/fac01/fac01.html"
But if I can avoid it, I try not to use it...
For problems like this, there's a trick so you don't have to use long division:$$\frac{u^2}{u^2 + 1} = \frac{u^2 + 1 - 1}{u^2 + 1} = \frac{u^2 + 1}{u^2 + 1} - \frac{1}{u^2 + 1} = 1 - \frac{1}{u^2 + 1}$$

 

[c.dube]

OK so now I have:

$2(\frac{1}{2}u^{2}tan^{-1}(u)-\frac{1}{2}(u-tan^{-1}(u))$

Or:

$(u^{2}+1)tan^{-1}(u)-u$

From here, I substitute sin(x) back in:

$(sin^{2}(x)+1)tan^{-1}(sin(x))-sin(x)$

Can I simplify more or am I still making a mistake? Thanks for all the help.

 

[Bohrok]

It's correct and probably as simplified as you can make it; just remember the + c

 

[c.dube]

Hahah yes that darn + C Thanks for all your help!