int frac{sin 2x}{sin^2 x + 3cos x - 3} , dx | vedclass

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(D) Let $I = \int \frac{\sin 2x}{\sin^2 x + 3\cos x - 3} \, dx$. Using $\sin 2x = 2\sin x \cos x$ and $\sin^2 x = 1 - \cos^2 x$,we get: $I = \int \fra

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$\log \left(\frac{(\cos x - 2)^4}{(\cos x - 1)^2}\right) + c$

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(D) Let $I = \int \frac{\sin 2x}{\sin^2 x + 3\cos x - 3} \, dx$. Using $\sin 2x = 2\sin x \cos x$ and $\sin^2 x = 1 - \cos^2 x$,we get: $I = \int \frac{2\sin x \cos x}{1 - \cos^2 x + 3\cos x - 3} \, dx = \int \frac{2\sin x \cos x}{-\cos^2 x + 3\cos x - 2} \, dx$. Let $t = \cos x$,then $dt = -\sin x \, dx$,so $\sin x \, dx = -dt$. $I = \int \frac{2t}{-t^2 + 3t - 2} (-dt) = \int \frac{2t}{t^2 - 3t + 2} \, dt$. Factor the denominator: $t^2 - 3t + 2 = (t - 1)(t - 2)$. Using partial fractions: $\frac{2t}{(t - 1)(t - 2)} = \frac{A}{t - 1} + \frac{B}{t - 2}$. $2t = A(t - 2) + B(t - 1)$. For $t = 1$,$2 = A(-1) \implies A = -2$. For $t = 2$,$4 = B(1) \implies B = 4$. $I = \int \left( \frac{-2}{t - 1} + \frac{4}{t - 2} \right) dt = -2 \log |t - 1| + 4 \log |t - 2| + c$. $I = \log |t - 2|^4 - \log |t - 1|^2 + c = \log \left| \frac{(t - 2)^4}{(t - 1)^2} \right| + c$. Substituting $t = \cos x$: $I = \log \left( \frac{(\cos x - 2)^4}{(\cos x - 1)^2} \right) + c$.

$\int \frac{\sin 2x}{\sin^2 x + 3\cos x - 3} \, dx$

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