Calculating the integral $\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$ I wanted to calculate $$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$
So I solved the indefinite integral first (by substitution):
$$\int\frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{b^2}\int\frac{d \theta}{\cos^2\theta \left(\frac{a^2}{b^2} \tan^2\theta+1 \right)} =\left[u=\frac{a}{b}\tan\theta, du=\frac{a}{b\cos^2\theta} d\theta \right ]\\=\frac{1}{b^2}\int\frac{b}{a\left(u^2+1 \right)}du=\frac{1}{ab}\int\frac{du}{u^2+1}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )+C$$
Then:
$$\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\frac{1}{ab} \arctan \left(\frac{a}{b}\tan (2\pi) \right )-\frac{1}{ab} \arctan \left(\frac{a}{b}\tan 0 \right )=0$$
Which is incorrect (the answer should be $2\pi/ab$ for $a>0,b>0$).
On the one hand, the substitution is correct, as well as the indefinite integral itself (according to Wolfram it is indeed $\frac{1}{ab} \arctan \left(\frac{a}{b}\tan\theta \right )$  ), but on the other hand I can see that had I put the limits during the substitution I'd get $\int\limits_{0}^{0} \dots = 0$ because for $\theta = 0 \to u=0$ and for $\theta = 2\pi \to u=0$.
Why is there a problem and how can I get the correct answer?
Edit: Here is Wolfram's answer:


Wolfram is correct because $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}$$ is the area of an ellipse (defined by $x=a\cos t , y=b\sin t$), that is $$\frac{a^2 b^2}{2}\int\limits_{0}^{2\pi} \frac{d \theta}{a^2 \sin^2\theta+b^2 \cos^2\theta}=\pi ab$$

You have everything right up to 
$$
\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))-\frac{1}{ab}\arctan(\frac{a}{b}\tan(0))
$$
 Now $\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))$ is $2\pi$ because the $\arctan$ and the $\tan$ are inverse functions.
So we get
$$
\frac{1}{ab}\arctan(\frac{a}{b}\tan(2\pi))-\frac{1}{ab}\arctan(\frac{a}{b}\tan(0))=\frac{1}{ab}2\pi-0
$$
or
$$
\frac{2\pi}{ab}
$$
