intlimits_0^{frac{pi }{2}} frac{dx}{1 + a^2 s | vedclass

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Question

(A) Let $I = \int\limits_0^{\frac{\pi}{2}} \frac{dx}{1 + a^2 \sin^2 x}$.Divide numerator and denominator by $\cos^2 x$:$I = \int\limits_0^{\frac{\pi}{

Vedclass · Accepted Answer

$\frac{\pi}{2\sqrt{1 + a^2}}$

Vedclass · Answer

(A) Let $I = \int\limits_0^{\frac{\pi}{2}} \frac{dx}{1 + a^2 \sin^2 x}$.Divide numerator and denominator by $\cos^2 x$:$I = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{\sec^2 x + a^2 	an^2 x} = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + 	an^2 x + a^2 	an^2 x} = \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 x \, dx}{1 + (1 + a^2) 	an^2 x}$.Let $u = 	an x$,then $du = \sec^2 x \, dx$. As $x 	o 0, u 	o 0$ and as $x 	o \frac{\pi}{2}, u 	o \infty$.$I = \int\limits_0^{\infty} \frac{du}{1 + (\sqrt{1 + a^2} u)^2}$.Using the formula $\int \frac{dx}{1 + k^2 x^2} = \frac{1}{k} 	an^{-1}(kx)$:$I = \left[ \frac{1}{\sqrt{1 + a^2}} 	an^{-1}(\sqrt{1 + a^2} u) ight]_0^{\infty} = \frac{1}{\sqrt{1 + a^2}} \left( \frac{\pi}{2} - 0 ight) = \frac{\pi}{2\sqrt{1 + a^2}}$.

$\int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 + a^2 \sin^2 x}$ has the value :

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