int frac{sin 2x}{4 sin^2 x + 9 cos^2 x} , dx | vedclass

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Question

(C) Let $I = \int \frac{\sin 2x}{4 \sin^2 x + 9 \cos^2 x} \, dx$. Let $t = 4 \sin^2 x + 9 \cos^2 x$. Differentiating both sides with respect to $x$: $

Vedclass · Accepted Answer

$-\frac{1}{5} \log(4 \sin^2 x + 9 \cos^2 x) + C$

Vedclass · Answer

(C) Let $I = \int \frac{\sin 2x}{4 \sin^2 x + 9 \cos^2 x} \, dx$. Let $t = 4 \sin^2 x + 9 \cos^2 x$. Differentiating both sides with respect to $x$: $dt = (4 \cdot 2 \sin x \cos x - 9 \cdot 2 \cos x \sin x) \, dx$. $dt = (4 \sin 2x - 9 \sin 2x) \, dx$. $dt = -5 \sin 2x \, dx$. Therefore,$\sin 2x \, dx = -\frac{1}{5} dt$. Substituting these into the integral: $I = \int \frac{-\frac{1}{5} dt}{t} = -\frac{1}{5} \int \frac{1}{t} \, dt$. $I = -\frac{1}{5} \log |t| + C$. Substituting back $t = 4 \sin^2 x + 9 \cos^2 x$: $I = -\frac{1}{5} \log |4 \sin^2 x + 9 \cos^2 x| + C$.

$\int \frac{\sin 2x}{4 \sin^2 x + 9 \cos^2 x} \, dx = $ (Where $C$ is a constant of integration).

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