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

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(C) To evaluate the integral $I = \int \frac{dx}{(2\sin x + \cos x)^2}$,we can rewrite the denominator by factoring out $\cos x$ from the expression i

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$\frac{1}{2 + \cot x} + c$

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(C) To evaluate the integral $I = \int \frac{dx}{(2\sin x + \cos x)^2}$,we can rewrite the denominator by factoring out $\cos x$ from the expression inside the square.$I = \int \frac{dx}{(\cos x (2	an x + 1))^2} = \int \frac{dx}{\cos^2 x (2	an x + 1)^2}$Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have:$I = \int \frac{\sec^2 x}{(2	an x + 1)^2} dx$Let $u = 2	an x + 1$. Then $du = 2\sec^2 x dx$,which implies $\sec^2 x dx = \frac{du}{2}$.Substituting these into the integral:$I = \int \frac{1}{u^2} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-2} du$$I = \frac{1}{2} \left( \frac{u^{-1}}{-1} ight) + c = -\frac{1}{2u} + c$Substituting $u = 2	an x + 1$ back:$I = -\frac{1}{2(2	an x + 1)} + c$Alternatively,using the method provided in the prompt:$I = \int \frac{dx}{\sin^2 x (2\csc x \sin x + \cot x)^2} = \int \frac{dx}{\sin^2 x (2 + \cot x)^2} = \int \frac{\csc^2 x}{(2 + \cot x)^2} dx$Let $t = 2 + \cot x$,then $dt = -\csc^2 x dx$,so $\csc^2 x dx = -dt$.$I = \int \frac{-dt}{t^2} = \frac{1}{t} + c = \frac{1}{2 + \cot x} + c$.

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

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