int frac{d x}{cos x sqrt{cos 2 x}} = | vedclass

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Question

(A) Let $I = \int \frac{d x}{\cos x \sqrt{\cos 2 x}}$.Divide the numerator and denominator by $\cos x$ inside the square root,or rewrite the expressio

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{d x}{\cos x \sqrt{\cos 2 x}}$.Divide the numerator and denominator by $\cos x$ inside the square root,or rewrite the expression:$I = \int \frac{d x}{\cos x \sqrt{\cos^2 x - \sin^2 x}} = \int \frac{d x}{\cos x \cdot \cos x \sqrt{1 - 	an^2 x}} = \int \frac{\sec^2 x}{\sqrt{1 - 	an^2 x}} d x$.Substitute $t = 	an x$,then $dt = \sec^2 x d x$.Thus,$I = \int \frac{dt}{\sqrt{1 - t^2}} = \sin^{-1}(t) + c$.Substituting back $t = 	an x$,we get $I = \sin^{-1}(	an x) + c$.

$\int \frac{d x}{\cos x \sqrt{\cos 2 x}} = $

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