Question

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 \(\alpha=\)
(A) \(\frac{2}{b^{2}}\)
(B) \(\frac{2}{a^{2}}\)
(C) \(-\frac{2}{b^{2}}\)
(D) \(-\frac{2}{a^{2}}\)

Answer 1

Ans(C)

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

Answer 2

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