# I Uncertain about an integral

1. Feb 10, 2017

### Mr Davis 97

I am trying to evaluate the integral $\displaystyle \int \frac{x}{1+\cos^2x}dx$. I have started by multiplying both the numerator and the denominator by $\frac{\sin^2x}{\cos^4x}$, to get $\displaystyle \int \frac{x\frac{\sin^2x}{\cos^4x}}{1+\tan^2x}dx$, and the denominator simplifies to sec^2, so we get $\displaystyle \int x \tan^2x~dx$. Is my reasoning up to this point correct?

2. Feb 10, 2017

### Staff: Mentor

I think you have a mistake in your work. If you multiply the denominator by $\frac{\sin^2x}{\cos^4x}$, you get $(1+\cos^2x)\frac{\sin^2x}{\cos^4x} = \frac{\sin^2x}{\cos^4x} + \tan^2x$. That first term on the right doesn't simplify to 1.

Wolframalpha gives a pretty complicated result: http://www.wolframalpha.com/input/?i=integrate+x/(1+++cos^2(x))+dx

3. Feb 12, 2017

### Svein

Using the complex domain: Substitute $u=e^{ix}$, then $x=\frac{1}{i}\log(u)= -i\log(u)$, $\frac{du}{dx}=ie^{ix}=iu$ and the integrand becomes
$$\frac{-i\log(u)}{1+(\frac{1}{2}(u+\frac{1}{u}))^{2}}=\frac{-i\log(u)}{1+(\frac{1}{4}(u^{2}+2+\frac{1}{u^{2}}))}=\frac{-4iu^{2}\log(u)}{4u^{2}+u^{4}+2u^{2}+1}$$
This can be rewritten as $$\frac{-2iu^{2}\log(u^{2})}{(u^{2}-3-2\sqrt{2})(u^{2}-3+2\sqrt{2})}$$ Now substitute z = u2.

Try to continue from there.