If int frac{sin 2 x}{(a+b cos x)^{2}} d x=alp | vedclass

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(C) Let $I = \int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$. Using the identity $\sin 2x = 2 \sin x \cos x$, we have $I = \int \frac{2 \sin x \cos x}{(a+

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$-\frac{2}{b^{2}}$

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(C) Let $I = \int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$. Using the identity $\sin 2x = 2 \sin x \cos x$, we have $I = \int \frac{2 \sin x \cos x}{(a+b \cos x)^{2}} d x$. Let $t = a + b \cos x$. Then $dt = -b \sin x \, dx$, which implies $\sin x \, dx = -\frac{dt}{b}$. Also, from $t = a + b \cos x$, we get $\cos x = \frac{t-a}{b}$. Substituting these into the integral: $I = \int \frac{2 (\frac{t-a}{b})}{t^2} \cdot (-\frac{dt}{b}) = -\frac{2}{b^2} \int \frac{t-a}{t^2} \, dt$. $I = -\frac{2}{b^2} \int (\frac{1}{t} - \frac{a}{t^2}) \, dt$. $I = -\frac{2}{b^2} [\ln |t| + \frac{a}{t}] + c$. Substituting $t = a + b \cos x$ back: $I = -\frac{2}{b^2} [\ln |a + b \cos x| + \frac{a}{a + b \cos x}] + c$. Comparing this with the given expression $\alpha [\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}]+c$, we find $\alpha = -\frac{2}{b^2}$.

If $\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x=\alpha\left[\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}\right]+c$, then $\alpha=$

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