int frac{1}{cos x sqrt{cos 2 x}} , dx = (whe | vedclass

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(A) Let $I = \int \frac{dx}{\cos x \sqrt{\cos 2x}}$.We know that $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$.So,$I = \int \frac{dx}{\cos x \sqrt{\frac{1

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

Vedclass · Answer

(A) Let $I = \int \frac{dx}{\cos x \sqrt{\cos 2x}}$.We know that $\cos 2x = \frac{1-	an^2 x}{1+	an^2 x}$.So,$I = \int \frac{dx}{\cos x \sqrt{\frac{1-	an^2 x}{1+	an^2 x}}} = \int \frac{dx}{\cos x \frac{\sqrt{1-	an^2 x}}{\sec x}}$.Since $\frac{1}{\cos x} = \sec x$,we have $I = \int \frac{\sec x \cdot \sec x}{\sqrt{1-	an^2 x}} \, dx = \int \frac{\sec^2 x}{\sqrt{1-	an^2 x}} \, dx$.Let $	an x = t$,then $\sec^2 x \, dx = dt$.Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{1-t^2}}$.Using the standard integral formula $\int \frac{dt}{\sqrt{1-t^2}} = \sin^{-1}(t) + C$,we get $I = \sin^{-1}(	an x) + C$.

$\int \frac{1}{\cos x \sqrt{\cos 2 x}} \, dx = $ (where $C$ is a constant of integration.)

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