int frac{sin 2x}{a^2 + b^2 sin^2 x} , dx = | vedclass

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Question

(A) Let $I = \int \frac{\sin 2x}{a^2 + b^2 \sin^2 x} \, dx$. Substitute $t = a^2 + b^2 \sin^2 x$. Differentiating with respect to $x$,we get $dt = b^2

Vedclass · Accepted Answer

$\frac{1}{b^2} \log(a^2 + b^2 \sin^2 x) + c$

Vedclass · Answer

(A) Let $I = \int \frac{\sin 2x}{a^2 + b^2 \sin^2 x} \, dx$. Substitute $t = a^2 + b^2 \sin^2 x$. Differentiating with respect to $x$,we get $dt = b^2 (2 \sin x \cos x) \, dx = b^2 \sin 2x \, dx$. Therefore,$\sin 2x \, dx = \frac{dt}{b^2}$. Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{dt}{b^2} = \frac{1}{b^2} \int \frac{1}{t} \, dt$. Integrating,we get $I = \frac{1}{b^2} \log |t| + c$. Substituting $t$ back,we get $I = \frac{1}{b^2} \log(a^2 + b^2 \sin^2 x) + c$.

$\int \frac{\sin 2x}{a^2 + b^2 \sin^2 x} \, dx = $

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