int frac{sec x , dx}{sqrt{cos 2x}} = | vedclass

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Question

(A) We have the integral $I = \int \frac{\sec x \, dx}{\sqrt{\cos 2x}}$.Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we get:$I = \int \frac{\sec

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$\sin^{-1}(	an x)$

Vedclass · Answer

(A) We have the integral $I = \int \frac{\sec x \, dx}{\sqrt{\cos 2x}}$.Using the identity $\cos 2x = \cos^2 x - \sin^2 x$,we get:$I = \int \frac{\sec x \, dx}{\sqrt{\cos^2 x - \sin^2 x}}$.Multiply the numerator and denominator by $\sec x$:$I = \int \frac{\sec^2 x \, dx}{\sqrt{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}}} = \int \frac{\sec^2 x \, dx}{\sqrt{1 - 	an^2 x}}$.Let $t = 	an x$,then $dt = \sec^2 x \, dx$.Substituting these into the integral:$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + C = \sin^{-1}(	an x) + C$.

$\int \frac{\sec x \, dx}{\sqrt{\cos 2x}} = $

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