_0^ 2 1 a^2 ^2 x+b^2 ^2 x dx is equal to - Vidyadip

MCQ

$\int_0^{\frac{\pi}{2}} \frac{1}{a^2 \cdot \sin ^2 x+b^2 \cdot \cos ^2 x}$ dx is equal to

A
$\frac{\pi a }{4 b}$
B
$\frac{\pi a }{2 b}$
C
$\frac{\pi b}{4 a}$
✓
$\frac{\pi}{2 a b}$

✓

Answer

Correct option: D.

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

(D)
Let $I =\int_0^{\frac{\pi}{2}} \frac{1}{ a ^2 \cdot \sin ^2 x+ b ^2 \cdot \cos ^2 x} d x$
$=\int_0^{\frac{\pi}{2}} \frac{1}{\cos ^2 x\left( a ^2 \tan ^2 x+ b ^2\right)} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{b^2+ a ^2 \tan ^2 x} d x$
Put a $\tan x= t \Rightarrow a ^{2 e c} x d x= dt$
$\therefore \quad I=\frac{1}{a} \int_0^{\infty} \frac{d t}{b^2+t^2}$
$=\frac{1}{ ab }\left[\tan ^{-1}\left(\frac{ t }{ b }\right)\right]_0^{\infty}=\frac{\pi}{2 ab }$

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