$\int_{0}^{π} \frac{x d x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}, w h e n a = b i s :$

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$\frac{π}{a b}$

$\frac{π^{2}}{a b}$

$\frac{2 π^{2}}{a b}$

$\frac{π^{2}}{2 a b}$

answer is D.

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Detailed Solution

$I = \int_{0}^{π} \frac{x d x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}$

$2 I = π \int_{0}^{π} \frac{d x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}$

$2 I = 2 π \int_{0}^{\frac{π}{2}} \frac{d x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}$

$I = 2 π \int_{0}^{\frac{π}{4}} \frac{d x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x}$

$= 2 π \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x d x}{a^{2} + b^{2} \tan^{2} x}$

$= 2 π \int_{0}^{1} \frac{d t}{a^{2} + b^{2} t^{2}}$ where $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$= \frac{2 π}{b^{2}} \times \frac{b}{a} \tan^{- 1} {(\frac{b t}{a})}_{0}^{1}$

$= \frac{2 π}{a b} \times \tan^{- 1} {(t)}_{0}^{1}$ , since $a = b$

$= \frac{2 π}{a b} \times \frac{π}{4} = \frac{π^{2}}{2 a b}$

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