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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 $$ \int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m} $$

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    $\begingroup$ What substitution(s) have you tried? $\endgroup$ – Cody Rudisill Apr 27 '16 at 18:59
  • $\begingroup$ I have try u = sinx $\endgroup$ – user334593 Apr 27 '16 at 19:03
  • $\begingroup$ Use the same way to get the general result. Should not be hard. $\endgroup$ – xpaul Apr 30 '16 at 13:29
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Let $$ I_1=\int_0^\pi \frac{dx}{\sqrt{n^2+1}+\sin(x)+n\cos(x)}, I_2=\int_0^\pi \frac{n dx}{\sqrt{n^2+1}+n\sin(x)+\cos(x)}. $$ Let $u=\tan\frac{x}{2}$. Then $x=2\arctan u$, $\sin x=\frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}$. Thus

\begin{eqnarray} I_1&=&\int_0^\infty \frac{1}{\sqrt{n^2+1}+\frac{2u}{1+u^2}+n\frac{1-u^2}{1+u^2}}\frac2{1+u^2}du\\ &=&2\int_0^\infty \frac{1}{\sqrt{n^2+1}(1+u^2)+2u+n(1-u^2)}du\\ &=&2\int_0^\infty \frac{1}{(\sqrt{n^2+1}-n)u^2+2u+(\sqrt{n^2+1}+n)}dx\\ &=&2\int_0^\infty \frac{\sqrt{n^2+1}+n}{u^2+2(\sqrt{n^2+1}+n)u+(\sqrt{n^2+1}+n)^2}du\\ &=&2\int_0^\infty \frac{\sqrt{n^2+1}+n}{(u+\sqrt{n^2+1}+n)^2}du\\ &=&2. \end{eqnarray} Similarly $$ I_2=2. $$ Done.

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First of all: $$ (\sqrt{n^2+1}+\sin x+n\cos x)(\sqrt{n^2+1}-\sin x-n\cos x)=$$ $$ =n^2+1-n^2\cos^2x-2n\cos x\sin x-\sin^2 x= $$ $$ =\cos^2x-2n\cos x\sin x+n^2\sin^2 x=(n\sin x-\cos x)^2 $$ So: $$ \int\frac{\mathrm{d}x}{\sqrt{n^2+1}+\sin x+n\cos x}=\int\frac{(\sqrt{n^2+1}-\sin x-n\cos x)\mathrm{d}x}{(n\sin x-\cos x)^2} $$ Now use the formula: $$ \theta=\arctan\frac{b}{a}\qquad a\cos x+b\sin x=\sqrt{a^2+b^2}{\cos(x+\theta)} $$ To finish this up.

Result: $$ \int\frac{\mathrm{d}x}{\sqrt{n^2+1}+\sin x+n\cos x}=\frac{1-\sqrt{n^2+1} \sin (x)}{n \sin (x)-\cos (x)}$$ $$\int\frac{n\mathrm{d}x}{\sqrt{n^2+1}+n\sin x+\cos x}=\frac{\sqrt{n^2+1} \sin (x)-n}{n \cos (x)-\sin (x)}$$

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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 $$\begin{align}\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}+n\sin x+\cos x}&=\int_0^{\pi}\frac{n\,dx}{\sqrt{1+n^2}(1+\cos(x-\phi))}\\ &=\int_{-\phi}^{\pi-\phi}\frac{n\,dy}{\sqrt{1+n^2}(1+\cos y)}\\ &=2\end{align}$$

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