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I want to evaluate

$$ \int_0^{\pi/2} \frac{a}{a^2+\cos^2 \theta} \, d\theta $$

and here is what Wolfram alpha gave me:

$$ \int_0^{\pi/2} \frac{a}{a^2+\cos^2 \theta} \, d\theta=\frac{\tan^{-1} \left(\frac{a \tan x}{\sqrt{a^2+1}}\right)}{\sqrt{1+a^2}} $$

Seeing the answer, I substituted $\cos \theta = \tan x$ hoping something good would happen, but in the end it didn't lead anywhere, assuming I didn't make any mistake. But after that I'm knid of hopelessly lost.

What's the magic trick here?

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$$\int_0^{\pi/2} \frac{a}{a^2+\cos^2 \theta} d\theta=\int_0^{\pi/2} \frac{a\sec^2 \theta}{a^2\sec^2 \theta + 1} d\theta$$ $$=\int_0^{\pi/2} \frac{a\sec^2 \theta}{a^2(1 + \tan^2 \theta) + 1} d\theta=\int_0^{\pi/2} \frac{a\sec^2 \theta}{a^2 + 1 + a^2 \tan^2 \theta} d\theta$$ Put $a\tan \theta = x$, $ a \sec^2 \theta d\theta = dx$, $$=\int_0^{\infty} \frac{1}{a^2 + 1 + x^2} dx =\frac{1}{\sqrt{a^2 + 1}}\arctan \frac{x}{\sqrt{a^2 + 1}} \biggl|_0^{\infty} =\frac{\pi}{2\sqrt{a^2 + 1}}$$

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  • $\begingroup$ Nice and simple. Thanks a lot! $\endgroup$ – user160738 Mar 11 '15 at 18:30
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Start using $\cos(2\theta)=2\cos ^2(\theta)-1$; so $$I=\int \frac{a}{a^2+\cos^2 (\theta)} \, d\theta=\int \frac{2a}{2a^2+1+\cos(2 \theta)} \, d\theta$$ Now, use the tangent half-angle substitution $\tan(\theta)=t$. So $$I=\int\frac{2 a}{a^2 \left(t^2+1\right)+1}\,dt=\int\frac{2 a}{a^2t^2+(a^2 +1)}\,dt=\frac2a\int\frac{dt}{t^2+\frac{a^2+1}{a^2}}$$ Now, $t=\sqrt{\frac{a^2+1}{a^2}}x$.

I am sure that you can take from here.

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  • $\begingroup$ So, my solution from wolfram: $\int _0^{\frac{\pi }{2}}\frac{a \text{d$\theta $}}{a^2+\cos ^2(\theta )}=\frac{\pi a}{2 \sqrt{a^2} \sqrt{a^2+1}}$ $\endgroup$ – Frieder Mar 11 '15 at 11:21
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Notice that: $$ \frac{a}{a^2 + cos^2\theta} = \frac{1}{2}(\frac{1}{a-icos\theta}+\frac{1}{a+icos\theta})$$ So it is enough to integrate on each of these summands. From here you can finish the calculation quite easily with the "Tangent half-angle substitution", i.e, $u = tan\frac{\theta}{2}$.

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  • $\begingroup$ Latex tips: Use '\cos' instead of 'cos'. To get large brackets use '\left( ... \right)' $\endgroup$ – Winther Mar 11 '15 at 22:51
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For the sake of completeness I present another common technique. Suppose we are trying to evaluate $$\int_0^{\pi/2} \frac{a}{a^2+\cos^2\theta} \; d\theta = \frac{1}{4} \int_0^{2\pi} \frac{a}{a^2+\cos^2\theta} \; d\theta.$$

Introduce $z=\exp(i\theta)$ so that $dz=iz\;d\theta$ to get $$\frac{1}{4} \int_{|z|=1} \frac{a}{a^2+(z+1/z)^2/4} \frac{1}{iz} \; dz \\ = \frac{1}{4} \int_{|z|=1} \frac{az^2}{z^2a^2+(z^2+1)^2/4} \frac{1}{iz} \; dz \\ = \frac{a}{i} \int_{|z|=1} \frac{z}{4z^2a^2+(z^2+1)^2} \; dz \\ = \frac{a}{i} \int_{|z|=1} \frac{z}{z^4+(4a^2+2)z^2+1} \; dz.$$

Call the function $f(z).$ The poles are at $$\pm \sqrt{-2a^2-1 \pm 2a \sqrt{a^2+1}}$$ which is $$\pm \sqrt{-2a^2-1 \pm 2a^2 \sqrt{1+1/a^2}}$$

When $a>1$ we have $$\sqrt{1+1/a^2} = 1 + \frac{1}{2} 1/a^2 - \frac{1}{8} 1/a^4 + \frac{1}{16} 1/a^6 + \cdots$$ Therefore $$\rho_{1,2} = \pm \sqrt{-2a^2-1 + 2a^2 \sqrt{1+1/a^2}} = \pm \sqrt{-\frac{1}{8} 1/a^4 + \frac{1}{16} 1/a^6 + \cdots}$$ so that these two poles are inside the unit circle.

On the other hand, $$\rho_{3,4} = \pm \sqrt{-2a^2-1 - 2a^2 \sqrt{1+1/a^2}} = \pm \sqrt{-4a^2 - 2 + \cdots}$$ so that these two poles are outside the unit circle.

It follows that the integral is given by $$\frac{a}{i} \times 2\pi i \times (\mathrm{Res}_{z=\rho_1} f(z) + \mathrm{Res}_{z=\rho_2} f(z))$$ or $$2a\pi \times (\mathrm{Res}_{z=\rho_1} f(z) + \mathrm{Res}_{z=\rho_2} f(z)).$$

These poles are simple so we get $$\mathrm{Res}_{z=\rho_{1,2}} f(z) =\rho_{1,2} \frac{1}{4\rho_{1,2}^3+2(4a^2+2)\rho_{1,2}} =\frac{1}{4\rho_{1,2}^2+2(4a^2+2)}.$$

This is $$\frac{1}{4(-2a^2-1)+ 8 a^2\sqrt{1+1/a^2} + 2(4a^2+2)}$$ or $$\frac{1}{8 a^2\sqrt{1+1/a^2}} = \frac{1}{8a \sqrt{a^2+1}}.$$

This finally yields for the integral $$2a\pi \times 2 \times \frac{1}{8a \sqrt{a^2+1}} = \frac{\pi}{2\sqrt{a^2+1}}.$$

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  • $\begingroup$ Nice! I've thought of doing this way, but I realized that things were going much too complicated after some steps that I was confident there are better ways to do this. $+1$ for your efforts. I was too lazy to do this all $\endgroup$ – user160738 Mar 11 '15 at 22:40
  • $\begingroup$ Thanks. Emphasis in comments is achieved by using enclosing pairs of stars. $\endgroup$ – Marko Riedel Mar 11 '15 at 22:42
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$$\small\int_0^{\pi/2} \frac{a}{a^2+\cos^2 \theta} {\rm d}\theta\stackrel{x=\tan\theta}=\int_0^{\infty} \frac{1}{1+a^2+(ax)^2} {\rm d}(ax)=\frac1{\sqrt{1+a^2}}\arctan\left(\frac{ax}{\sqrt{1+a^2}}\right)\Bigg|_{0}^{\infty}=\frac{\pi}{2\sqrt{1+a^2}}$$

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