int frac{sin 2x , dx}{1 + cos^2 x} = | vedclass

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Question

(A) Let $I = \int \frac{\sin 2x}{1 + \cos^2 x} \, dx$. Using the identity $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{2 \sin x \cos x}{1 + \c

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{\sin 2x}{1 + \cos^2 x} \, dx$. Using the identity $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{2 \sin x \cos x}{1 + \cos^2 x} \, dx$. 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: $I = \int \frac{-dt}{t} = -\log|t| + c$. Replacing $t$ with $1 + \cos^2 x$: $I = -\log(1 + \cos^2 x) + c$.

$\int \frac{\sin 2x \, dx}{1 + \cos^2 x} = $

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