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

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(D) Let $I = \int \frac{\sin 2x}{\sin^2 x + 2\cos^2 x} dx$. We know that $\sin^2 x + \cos^2 x = 1$,so the denominator becomes $\sin^2 x + \cos^2 x + \

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

Vedclass · Answer

(D) Let $I = \int \frac{\sin 2x}{\sin^2 x + 2\cos^2 x} dx$. We know that $\sin^2 x + \cos^2 x = 1$,so the denominator becomes $\sin^2 x + \cos^2 x + \cos^2 x = 1 + \cos^2 x$. Thus,$I = \int \frac{\sin 2x}{1 + \cos^2 x} dx$. Let $t = 1 + \cos^2 x$. Then $dt = 2\cos x(-\sin x) dx = -\sin 2x dx$. So,$\sin 2x dx = -dt$. Substituting these into the integral,we get $I = \int \frac{-dt}{t} = -\log|t| + C$. Substituting back $t = 1 + \cos^2 x$,we get $I = -\log(1 + \cos^2 x) + C$.

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

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