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

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(B) We are given the integral $I = \int \frac{dx}{\cos 2x + \sin^2 x}$.Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we substitute this i

Vedclass · Accepted Answer

$	an x + c$

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(B) We are given the integral $I = \int \frac{dx}{\cos 2x + \sin^2 x}$.Using the trigonometric identity $\cos 2x = 1 - 2\sin^2 x$,we substitute this into the denominator:$I = \int \frac{dx}{1 - 2\sin^2 x + \sin^2 x}$$I = \int \frac{dx}{1 - \sin^2 x}$Using the identity $1 - \sin^2 x = \cos^2 x$,we get:$I = \int \frac{dx}{\cos^2 x}$$I = \int \sec^2 x \, dx$The integral of $\sec^2 x$ is $	an x + c$.Therefore,$I = 	an x + c$.

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

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