Question

Evaluate ${\int }_0^{\pi /2}\frac{{\cos }^{2}x}{1+3{\sin }^{2}x}dx$.

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{9}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (D)

$\frac{\pi }{6}$

Let $I={\int }_0^{\pi /2}\frac{{\cos }^{2}x}{1+3{\sin }^{2}x}dx$. Then

$I={\int }_0^{\pi /2}\frac{{\cos }^{2}x}{1+3{\sin }^{2}x}dx$. Then,

$I={\int }_0^{\pi /2}\frac{{\sec }^{2}x}{{\sec }^{2}x({\sec }^{2}x+3{\tan }^{2}x)}dx$ [Diving $N^r$ and $D^r$ by ${\cos }^{4}x$]

$\Rightarrow I={\int }_0^{\infty }\frac{1}{(1+t^2)(1+4t^2)}dt$, where $t=\tan x$

$\Rightarrow I=-\frac{1}{3}{\int }_0^{\infty }(\frac{1}{1+t^2}-\frac{4}{1+t^2})dt=-\frac{1}{3}{[{\tan }^{-1}t-2{\tan }^{-1}2t]}_0^{\infty }$

$\Rightarrow =-\frac{1}{3}(\frac{\pi }{2}-\pi )=\frac{\pi }{6}$