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

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(B) Let $I = \int \frac{\sin 2x}{(a+b \cos x)^2} dx = \int \frac{2 \sin x \cos x}{(a+b \cos x)^2} dx$. Substitute $t = a + b \cos x$,then $dt = -b \si

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$\frac{-2}{b^2} \left[ \log |a+b \cos x| + \frac{a}{a+b \cos x} \right] + C$

Vedclass · Answer

(B) Let $I = \int \frac{\sin 2x}{(a+b \cos x)^2} dx = \int \frac{2 \sin x \cos x}{(a+b \cos x)^2} dx$. Substitute $t = a + b \cos x$,then $dt = -b \sin x dx$,so $\sin x dx = -\frac{dt}{b}$. Also,$\cos x = \frac{t-a}{b}$. Substituting these into the integral: $I = \int \frac{2 (\frac{t-a}{b})}{t^2} (-\frac{dt}{b}) = -\frac{2}{b^2} \int \frac{t-a}{t^2} dt$. $I = -\frac{2}{b^2} \int (\frac{1}{t} - \frac{a}{t^2}) dt$. $I = -\frac{2}{b^2} [\log |t| + \frac{a}{t}] + C$. Substituting $t = a + b \cos x$ back: $I = -\frac{2}{b^2} [\log |a+b \cos x| + \frac{a}{a+b \cos x}] + C$.

$\int \frac{\sin 2x}{(a+b \cos x)^2} dx =$

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