Question

The value of the integral ${\int }_0^{2\pi }\left(\frac{3}{9+si{n}^2\theta }\right)d\theta$ is

• A. $\frac{2\pi }{\sqrt{10}}$
• B. $2\sqrt{10}\pi$
• C. $\sqrt{10}\pi$
• D. $2\pi$

Answer 1

The correct option is A $\frac{2\pi }{\sqrt{10}}$

$I={\int }_0^{2\pi }\frac{3}{9+si{n}^2\theta }d\theta =2{\int }_0^{\pi }\frac{3}{9+si{n}^2\theta }d\theta$
$=4{\int }_0^{\pi /2}\frac{3d\theta }{9+si{n}^2\theta }$
$=12{\int }_0^{\pi /2}\frac{se{c}^2\theta d\theta }{9se{c}^2\theta +ta{n}^2\theta }$
$=12{\int }_0^{\pi /2}\frac{se{c}^2\theta d\theta }{9+10ta{n}^2\theta }$
Put $\sqrt{10}tan\theta =t\Rightarrow se{c}^2\theta d\theta =\frac{dt}{\sqrt{10}}$
So, $I=12{\int }_0^{\infty }\frac{dt/\sqrt{10}}{9+t^2}=\frac{12}{\sqrt{10}}{\left[\frac{1}{3}ta{n}^{-1}\left(\frac{t}{3}\right)\right]}_0^{\infty }$
$=\frac{4}{\sqrt{10}}\left[ta{n}^{-1}\right(\infty )-ta{n}^{-1}(0\left)\right]=\frac{2\pi }{\sqrt{10}}$