Trigonometric Integral

Evaluate the following:

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

#Calculus #Trigonometry #Integration #DefiniteIntegral #JEE

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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

Haroun Meghaichi - 6 years, 11 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:

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

Pranav Arora - 6 years, 11 months ago

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

Haroun Meghaichi - 6 years, 11 months ago

@Haroun Meghaichi – Try this one too: https://brilliant.org/discussions/thread/an-interesting-double-definite-integral/?ref_id=294562. :)

Pranav Arora - 6 years, 11 months ago

0.916298

Rajdeep Ghosh - 6 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$