TITLE: How do I go about solving this?
QUESTION [6 upvotes]: I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$

REPLY [2 votes]: This seems so easy in polar form. Let $1=r\cos\phi$, $n=r\sin\phi$, then $r=\sqrt{1+n^2}$, $\phi=\tan^{-1}n$ and then let $x+\phi=y+\frac{\pi}2$
$$\begin{align}\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}+\sin x+n\cos x}&=\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}(1+\sin(x+\phi))}\\
&=\int_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\frac{dy}{\sqrt{1+n^2}(1+\cos y)}\\
&=\left.\frac{\sin y}{\sqrt{1+n^2}(1+\cos y)}\right|_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\\
&=\frac1{\sqrt{1+n^2}}\left(\frac{\cos\phi}{1-\sin\phi}+\frac{\cos\phi}{1+\sin\phi}\right)\\
&=\frac1{\sqrt{1+n^2}}\left(\frac1{\sqrt{1+n^2}-n}+\frac1{\sqrt{1+n^2}+n}\right)\\
&=2\end{align}$$
The second integral, after the substitution $y=x-\phi$ becomes
$$\begin{align}\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}+n\sin x+\cos x}&=\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}(1+\cos(x-\phi))}\\
&=\int_{-\phi}^{\pi-\phi}\frac{n\,dy}{\sqrt{1+n^2}(1+\cos y)}\\
&=2\end{align}$$