Question

$\stackrel{\pi /2}{\underset{0}{\int }}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ equals-

• A. $\pi /ab$
• B. $2\pi /ab$
• C. $ab/\pi$
• D. $\pi /2ab$

Answer 1

The correct option is A $\pi /ab$

$I={\int }_0^{\pi /2}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$={\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$

Let $\tan x=t$ $x=0,\tan x=0$

${\sec }^{2}xdx=dt$ $x=\pi /2,\tan x=\infty$

$\Rightarrow {\int }_0^{\infty }\frac{dt}{a^2+b^2{t}^2}=\frac{1}{b^2}{\int }_0^{\infty }\frac{dt}{(\frac{a^2}{b^2}+t^2)}$

$\Rightarrow \frac{1}{b^2}\times \frac{1}{\left(a/b\right)}{\left[{\tan }^{-1}\left(\frac{t}{\left(a/b\right)}\right)\right]}_0^{\infty }$

$=\frac{1}{ab}\left[{\tan }^{-1}\right(\infty )-{\tan }^{-1}(0\left)\right]$

$=\frac{1}{ab}\left[\pi /2\right]$

$\Rightarrow \frac{\pi }{2ab}$.