Question

What is ${\int }_0^{\frac{\pi }{2}}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ equal to?

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

Answer 1

The correct option is C $\frac{\pi }{2ab}$

Given $\int \frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$=\int \frac{dx}{a^2{\cos }^{2}x(1+\frac{b^2}{a^2}{\tan }^{2}x)}$

$=\frac{1}{a^2}\int \frac{{\sec }^{2}xdx}{(1+\frac{b^2}{a^2}{\tan }^{2}x)}$

let $u=\frac{b}{a}\tan x\Rightarrow du=\frac{b}{a}{\sec }^{2}xdx.................EQ\left(1\right)$

Changing the limits accordingly $u=\infty andu=0$

Substituting $EQ\left(1\right)$ in the integral

$=\frac{1}{a^2}{\int }_0^{\infty }\frac{\frac{a}{b}du}{1+u^2}$

$=\frac{1}{ab}\int \frac{du}{1+u^2}$

$={\left[\frac{1}{ab}{\tan }^{-1}x\right]}_0^{\infty }$

$=\frac{1}{ab}({\tan }^{-1}\infty -{\tan }^{-1}0)$

$=\frac{1}{ab}(\frac{\pi }{2}-0)$

$=\frac{\pi }{2ab}$