int frac{1}{(2 cos x+sin x)^2} d x= | vedclass

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Question

(D) Let $I = \int \frac{1}{(2 \cos x+\sin x)^2} d x$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{(2 + 	an x)^2} d

Vedclass · Accepted Answer

$-\frac{\cos x}{2 \cos x+\sin x}+c$

Vedclass · Answer

(D) Let $I = \int \frac{1}{(2 \cos x+\sin x)^2} d x$. Divide the numerator and denominator by $\cos^2 x$: $I = \int \frac{\sec^2 x}{(2 + 	an x)^2} d x$. Let $u = 2 + 	an x$. Then $du = \sec^2 x \, dx$. Substituting these into the integral: $I = \int \frac{1}{u^2} du = \int u^{-2} du$. Integrating with respect to $u$: $I = \frac{u^{-1}}{-1} + c = -\frac{1}{u} + c$. Substituting back $u = 2 + 	an x$: $I = -\frac{1}{2 + 	an x} + c$. Since $	an x = \frac{\sin x}{\cos x}$, we have: $I = -\frac{1}{2 + \frac{\sin x}{\cos x}} + c = -\frac{\cos x}{2 \cos x + \sin x} + c$. Thus, the correct option is $D$.

$\int \frac{1}{(2 \cos x+\sin x)^2} d x=$

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