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

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(D) Let $I = \int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx$. Using the identity $\sin^2(x) = 1 - \cos^2(x)$,the denominator becomes $1 - \cos^2(x) +

Vedclass · Accepted Answer

$-\log |1 + \cos^2(x)| + c$

Vedclass · Answer

(D) Let $I = \int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx$. Using the identity $\sin^2(x) = 1 - \cos^2(x)$,the denominator becomes $1 - \cos^2(x) + 2\cos^2(x) = 1 + \cos^2(x)$. So,$I = \int \frac{\sin(2x)}{1 + \cos^2(x)} dx$. Let $t = 1 + \cos^2(x)$. Then $dt = 2\cos(x)(-\sin(x)) dx = -\sin(2x) dx$. This implies $\sin(2x) dx = -dt$. Substituting these into the integral: $I = \int \frac{-dt}{t} = -\log |t| + c$. Substituting back $t = 1 + \cos^2(x)$,we get $I = -\log |1 + \cos^2(x)| + c$. Thus,option $D$ is correct.

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

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