int frac{1}{1 + sin^2 x} , dx = | vedclass

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Question

(A) Let $I = \int \frac{1}{1 + \sin^2 x} \, dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{\sec^2 x + 	an^2 x} \,

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} 	an^{-1}(\sqrt{2} 	an x) + k$

Vedclass · Answer

(A) Let $I = \int \frac{1}{1 + \sin^2 x} \, dx$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{\sec^2 x + 	an^2 x} \, dx = \int \frac{\sec^2 x}{1 + 	an^2 x + 	an^2 x} \, dx = \int \frac{\sec^2 x}{1 + 2	an^2 x} \, dx$. Substitute $t = 	an x$,so $dt = \sec^2 x \, dx$: $I = \int \frac{dt}{1 + 2t^2} = \frac{1}{2} \int \frac{dt}{\frac{1}{2} + t^2} = \frac{1}{2} \int \frac{dt}{(\frac{1}{\sqrt{2}})^2 + t^2}$. Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$: $I = \frac{1}{2} \cdot \frac{1}{1/\sqrt{2}} 	an^{-1}(\frac{t}{1/\sqrt{2}}) + k = \frac{\sqrt{2}}{2} 	an^{-1}(\sqrt{2}t) + k = \frac{1}{\sqrt{2}} 	an^{-1}(\sqrt{2} 	an x) + k$.

$\int \frac{1}{1 + \sin^2 x} \, dx = $

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