Question

If $I={\int }_0^{\pi }\frac{dx}{5+3\cos x}$ then $I$ equals

• A. $\pi$
• B. $2\pi /3$
• C. $\pi /4$
• D. $2\pi /\sqrt{3}$

Answer 1

The correct option is C $\pi /4$

Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx,$
We can write
$I={\int }_0^{\pi }\frac{dx}{5+3\cos (\pi -x)}={\int }_0^{\pi }\frac{dx}{5-3\cos x}\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right),$ we get

$2I={\int }_0^{\pi }[\frac{1}{5+3\cos x}+\frac{1}{5-3\cos x}]dx$
$I=5{\int }_0^{\pi }\left[\frac{dx}{25-9{\cos }^{2}x}\right]$
$I=5{\int }_0^{\pi }\left[\frac{dx}{16+9{\sin }^{2}x}\right]=5{\int }_0^{\pi }\left[\frac{cose{c}^{2}x}{16{\cot }^{2}x+25}\right]dx$
$=\frac{1}{4}{\tan }^{-1}{\left(\frac{4\cot x}{5}\right)]}_0^{\pi }$

$=-\frac{1}{4}\{-\frac{\pi }{2}-\frac{\pi }{2}\}=\frac{\pi }{4}$