_0^ / 2 d x a^2 ^2 x+b^2 ^2 x = - Vidyadip

MCQ

$\int_0^{\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}=$

A
$\pi ab$
B
$\pi^2 a b$
C
$\frac{\pi}{a b}$
✓
$\frac{\pi}{2 a b}$

✓

Answer

Correct option: D.

$\frac{\pi}{2 a b}$

(D)
Let $I =\int_0^{\pi / 2} \frac{d x}{ a ^2 \cos ^2 x+ b ^2 \sin ^2 x}$
Dividing numerator and denominator by $\cos ^2 x$,
we get
$I =\int_0^{\pi / 2} \frac{\sec ^2 x}{ a ^2+ b ^2 \tan ^2 x} d x$
Put $b \tan x= t \Rightarrow b \ sec ^2 \ x \ d x= dt$
When $x=0, t =0$ and when $x=\frac{\pi}{2}, t =\infty$
$\therefore I=\int_0^{\infty} \frac{\frac{ dt }{ b }}{ a ^2+ t ^2}=\frac{1}{b}\left[\frac{1}{ a } \tan ^{-1}\left(\frac{ t }{ a }\right)\right]_0^{\infty}$
$=\frac{1}{ ab }\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)$
$=\frac{1}{ ab }\left(\frac{\pi}{2}-0\right)=\frac{\pi}{2 ab }$

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