Q83 / 155 · WBJEE 2021 · Mathematics

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
    $\frac{2}{\mathrm{~b}^{2}}$
  • B
    $\frac{2}{a^{2}}$
  • Correct answer
    C
    $-\frac{2}{b^{2}}$
  • D
    $-\frac{2}{\mathrm{a}^{2}}$

Solution

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

Hence, the correct answer is option: 'C'.

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