\(\tan \frac{x}{2} =t\) is a killer in these cases, the proposed integral equals : \[\int_0^{\infty}\left(\frac{ \frac{1-t^2}{t^2+1}+2}{\frac{4 \left(1-t^2\right)}{t^2+1}+5 }\right)^2 \frac{2}{t^2+1} \ \mathrm{d}t= \int_0^{\infty} \frac{15}{8 \left(t^2+9\right)}-\frac{9}{\left(t^2+9\right)^2}+\frac{1}{8 \left(t^2+1\right)} \ \mathrm{d}t \] The latter integral is quite easy (if I'm not mistaken it is \(\frac{7\pi}{24}\) ).

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TopNewest\(\tan \frac{x}{2} =t\) is a killer in these cases, the proposed integral equals : \[\int_0^{\infty}\left(\frac{ \frac{1-t^2}{t^2+1}+2}{\frac{4 \left(1-t^2\right)}{t^2+1}+5 }\right)^2 \frac{2}{t^2+1} \ \mathrm{d}t= \int_0^{\infty} \frac{15}{8 \left(t^2+9\right)}-\frac{9}{\left(t^2+9\right)^2}+\frac{1}{8 \left(t^2+1\right)} \ \mathrm{d}t \] The latter integral is quite easy (if I'm not mistaken it is \(\frac{7\pi}{24}\) ).

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Nice! Another way to go about it is to recognise:

\(\displaystyle \frac{2+\cos x}{5+4\cos x}=\sum_{n=0}^{\infty} \frac{(-1)^n\cos (nx)}{2^{n+1}}\).

Hence,

\(\displaystyle \int_0^{\pi} \left(\frac{2+\cos x}{5+4\cos x}\right)^2\,dx=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{(-1)^{m+n}}{2^{m+n+2}}\int_0^{\pi}\cos(mx)\cos(nx)\,dx \)

The integral is zero unless \(n=m\), so the sum reduces to:

\(\displaystyle \sum_{n=0}^{\infty} \frac{1}{2^{2n+2}}\int_0^{\pi} \cos^2(nx)\,dx=\frac{\pi}{4}+\frac{\pi}{8}\sum_{n=1}^{\infty} \frac{1}{2^{2n}}=\boxed{\dfrac{7\pi}{24}}\)

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Good, and the sum-integral interchange is justified by the dominated convergence theorem (in the Riemanian sense).

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https://brilliant.org/discussions/thread/an-interesting-double-definite-integral/?ref_id=294562. :)

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0.916298

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