<s>[INST] Definite integral of $1/(5+4\cos x)$ over $2$ periods Question:
$$\int_0^{4\pi}\frac{dx}{5+4\cos x} $$
My approach: 
First I calculated the antiderivative as follows:  
Using: $\cos\theta= \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$ we have:
$\int\frac{dx}{5+4\cos x}=\int\frac{dx}{5+4\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}=\int\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}+4-4\tan^2\frac{x}{2}}dx=\int\frac{\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx$
Using substitution we have:  
$u=\tan\frac{x}{2}$
$du=\frac{1}{2}\frac{1}{\cos^2\frac{x}{2}}dx$ 
$2\int\frac{\frac{1}{2}\frac{1}{\cos^2 \frac{x}{2}}}{3^2+\tan^2\frac{x}{2}}dx=2\int\frac{du}{3^2+u^2}=\frac{2}{3}\arctan\frac{u}{3}+\mathscr{C}=\frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}+ \mathscr{C}$
Now we can calculate the definite integral as follows:  
$\int_0^{4\pi}\frac{dx}{5+4\cos x} = \frac{2}{3}\arctan\frac{\tan\frac{x}{2}}{3}\bigl|_0^{4\pi}=\frac{2}{3}(\arctan\frac{\tan\frac{4\pi}{2}}{3}-\arctan\frac{\tan\frac{0}{2}}{3})=0$ 
The result I get is $0$ but the correct one is $\frac{4\pi}{3}$. Can someone explain me why?  
Here it shows that the correct answer is $\frac{4\pi}{3}$.
 [/INST] Not an answer to the question but a quick note: we can clean up your computation by working in terms of $u$ rather than in terms of $x$. With the substitution of $u = \tan(x/2)$, we find that
$$
du=\frac{1}{2}\sec^2\frac{x}{2}dx = \frac 12 (1 + u^2)\,dx
$$
Now, we have
$$
\int \frac{1}{5 + 4\cos x}dx =
\int \frac{1}{5 + 4\frac{1-u^2}{1+u^2}}dx =
\int \frac{(1+u^2)}{5(1+u^2) + 4(1-u^2)}dx =
\int \frac{(1+[u(x)]^2)}{3^2 + [u(x)]^2}\,dx.
$$
From here, substitution gives us
$$
2\int \frac{1}{3^2 + [u(x)]^2}\cdot\frac{1+[u(x)]^2}{2} dx = 2\int\frac{1}{3^2 + u^2}\,du.
$$
</s>