How do I evaluate this integral by hand? TL;DR how do I evaluate $\int_0^{2 \pi } \frac{1}{\cos ^2(\theta )+1} \, d\theta$ by hand?
I'm trying to solve this problem:

Find the volume of the region defined by $x^2+xy+y^2+yz+z^2\le1$.
(The answer is $\frac{4 \sqrt{2} \pi }{3}$)

I have tried several tactics. First of all, I rotated it $45^\circ$ in the $xz$ plane (converting all $x$ to $\frac{(x-z)}{\sqrt{2}}$ and all $z$ to $\frac{(x+z)}{\sqrt{2}}$ to obtain this:
$$x^2+y^2+z^2+xy\sqrt{2}\le1$$
Now if I solve for y at the boundary I get
$$y=\frac{1}{2} \left(-\sqrt{2} x\pm \sqrt{2} \sqrt{-x^2-2 z^2+2}\right)$$
And then I can integrate out the $y$ to get
$$\int _{-1}^1\int _{-\sqrt{2} \sqrt{1-z^2}}^{\sqrt{2} \sqrt{1-z^2}}\sqrt{2}
   \sqrt{-x^2-2 z^2+2}\ dxdz$$
Which I can evaluate with Mathematica but not by hand [it does get the right answer in Mathematica, so I know I'm on the right track.]
I can then switch to polar coordinates in $x$ and $z$, yielding 
$$\sqrt{2}  \int _0^{2 \pi }\int _0^{\sqrt{\frac{2}{2 \sin ^2(\theta )+\cos ^2(\theta
   )}}}r \sqrt{-2 r^2 \sin ^2(\theta )-r^2 \cos ^2(\theta )+2}drd\theta$$
This is relatively easy to integrate (the inner integral), so I get
$$\frac{4}{3} \int_0^{2 \pi } \frac{1}{\cos ^2(\theta )+1} \, d\theta$$
which I can't figure out how to evaluate by hand. Where do I go from here?

Here is an approach. First evaluate the indefinte integral using the substitution $\theta=\arctan(u)$ which gives

$$ I=\int \frac{d\theta}{1+\cos^2(\theta)} = \int \frac{du}{u^2+2} =\frac{1}{\sqrt{2}}\arctan\left(\frac{u}{\sqrt{2}}\right).$$

Then, split the interval of integration as $[0,\pi/2]\cup[\pi/2,\pi]\cup [\pi,3\pi/2]\cup[3\pi/2,2\pi]$. In fact, you can write the integral as

$$ I = 4\int_{0}^{\infty} \frac{du}{u^2+2}= 4 \frac{\sqrt{2}\pi}{4}=\sqrt{2}\pi. $$

Note:

$$ \cos^2(\tan^{-1}(u)) = \frac{1}{1+u^2}. $$

