int_{0}^{frac{pi}{2}} frac{1}{a^{2} sin ^{2} | vedclass

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(B) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $. Divide numerator and denominator by $ \cos^{2} x $: $ I =

Vedclass · Accepted Answer

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

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(B) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $. Divide numerator and denominator by $ \cos^{2} x $: $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^{2} x}{a^{2} 	an^{2} x + b^{2}} d x $. Let $ 	an x = t $,then $ \sec^{2} x d x = d t $. When $ x = 0, t = 0 $ and when $ x = \frac{\pi}{2}, t 	o \infty $. $ I = \int_{0}^{\infty} \frac{d t}{a^{2} t^{2} + b^{2}} = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{d t}{t^{2} + (b/a)^{2}} $. Using the formula $ \int \frac{d x}{x^{2} + k^{2}} = \frac{1}{k} 	an^{-1}(\frac{x}{k}) $: $ I = \frac{1}{a^{2}} \cdot \frac{1}{(b/a)} \left[ 	an^{-1} \left( \frac{t}{b/a} ight) ight]_{0}^{\infty} $. $ I = \frac{1}{ab} \left( 	an^{-1}(\infty) - 	an^{-1}(0) ight) = \frac{1}{ab} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{2ab} $.

$ \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $ is equal to

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