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Trigonometric Integral

Evaluate the following:

$\int_0^{\pi} \left(\frac{2+\cos x}{5+4\cos x}\right)^2\,dx$

Note by Pranav Arora
3 years, 8 months ago

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$$\tan \frac{x}{2} =t$$ 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 $$\frac{7\pi}{24}$$ ).

- 3 years, 8 months ago

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:

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

- 3 years, 8 months ago

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

- 3 years, 8 months ago

0.916298

- 3 years, 8 months ago