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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Aicharya · Accepted Answer

$\int \frac{\sin 2 x}{(a+b \cos x)^{2}} d x$ Put $a+b \cos x=t$ $-b \sin x d x=d t$ $=\int \frac{2 \sin x \cos x}{t^{2}} \cdot \frac{d t}{-b \sin x}=\int \frac{-2}{b} \cdot\left(\frac{t-a}{b}ight) \cdot \frac{1}{t^{2}} d t=\frac{-2}{b^{2}} \int\left(\frac{1}{t}-\frac{a}{t^{2}}ight) d t \quad 	herefore \alpha=\frac{-2}{b^{2}}$ $=-\frac{2}{b^{2}}\left[\ln |t|+\frac{a}{t}ight]+c$ $=\frac{-2}{b^{2}}\left[\ln |a+b \cos x|+\frac{a}{a+b \cos x}ight]+c$

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…

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