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

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Question

(B) Let $I = \int \frac{\sin 2x}{1 + \sin^2 x} dx$. Substitute $t = 1 + \sin^2 x$. Differentiating both sides with respect to $x$,we get $dt = \frac{d

Vedclass · Accepted Answer

$\log (1 + \sin^2 x) + c$

Vedclass · Answer

(B) Let $I = \int \frac{\sin 2x}{1 + \sin^2 x} dx$. Substitute $t = 1 + \sin^2 x$. Differentiating both sides with respect to $x$,we get $dt = \frac{d}{dx}(1 + \sin^2 x) dx = 2 \sin x \cos x dx = \sin 2x dx$. Substituting these into the integral,we get $I = \int \frac{1}{t} dt$. Integrating,we get $I = \log |t| + c$. Substituting back $t = 1 + \sin^2 x$,we get $I = \log (1 + \sin^2 x) + c$ (since $1 + \sin^2 x > 0$ for all $x$).

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

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