If int frac{sqrt{2} , dx}{cos x sqrt{sin 2x}} | vedclass

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(C) Given integral: $I = \int \frac{\sqrt{2} \, dx}{\cos x \sqrt{\sin 2x}}$.Using the identity $\sin 2x = 2 \sin x \cos x$, we have $\sqrt{\sin 2x} =

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$2 \sqrt{	an x}$

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(C) Given integral: $I = \int \frac{\sqrt{2} \, dx}{\cos x \sqrt{\sin 2x}}$.Using the identity $\sin 2x = 2 \sin x \cos x$, we have $\sqrt{\sin 2x} = \sqrt{2} \sqrt{\sin x} \sqrt{\cos x}$.Substituting this into the integral:$I = \int \frac{\sqrt{2} \, dx}{\cos x \cdot \sqrt{2} \sqrt{\sin x} \sqrt{\cos x}} = \int \frac{dx}{\cos x \sqrt{\sin x} \sqrt{\cos x}} = \int \frac{dx}{(\cos x)^{3/2} \sqrt{\sin x}}$.Divide numerator and denominator by $\cos^2 x$:$I = \int \frac{\sec^2 x \, dx}{\sqrt{\frac{\sin x}{\cos x}}} = \int \frac{\sec^2 x \, dx}{\sqrt{	an x}}$.Let $u = 	an x$, then $du = \sec^2 x \, dx$.$I = \int \frac{du}{\sqrt{u}} = \int u^{-1/2} \, du = 2u^{1/2} + c = 2 \sqrt{	an x} + c$.Thus, $f(x) = 2 \sqrt{	an x}$.

If $\int \frac{\sqrt{2} \, dx}{\cos x \sqrt{\sin 2x}} = f(x) + c$, then $f(x) =$

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