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

(A) $\dfrac{2}{\mathrm{~b}^{2}}$

(B) $\dfrac{2}{\mathrm{a}^{2}}$

(C) $-\dfrac{2}{\mathrm{~b}^{2}}$

(D) $-\dfrac{2}{\mathrm{a}^{2}}$

Take the $u$-substitution $\cos x=u$ then $-\sin x\ dx=du$. This leads us

\begin{aligned}\int \frac{\sin 2 x}{(a+b \cos x)^{2}} \ d x&= \int \frac{2 \cos x\sin x}{(a+b \cos x)^{2}} d x \\ &=-2\int \frac{u\ du}{(a+b u)^{2}} \end{aligned} \tag{Eq. 1}

We will also perform the same $u$-substitution in the RHS of the given expression to get

$$\alpha\left[\log _{e}|a+b \cos x|+\frac{a}{a+b \cos x}\right]+c=\alpha\left[\log _{e}|a+b u|+\frac{a}{a+b u}\right]+c\tag{Eq. 2}$$

By comparing (Eq. 1) and (Eq. 2) we get

$$-2\int \frac{u\ du}{(a+b u)^{2}}= \alpha\left[\log _{e}|a+b u|+\frac{a}{a+b u}\right]+c \tag{Eq. 3}$$

We will not evaluate the integration to remove integration sign from left we will differentiate both sides with respect to $u$ to obtain

\begin{aligned}\frac{-2u}{(a+b u)^{2}}&= \alpha\left[\frac{1}{a+bu}+\frac{0-ab}{(a+bu)^2}\right] \\ \frac{-2u}{(a+b u)^{2}}&= \alpha\left[\frac{a+bu-ab}{(a+bu)^2}\right] \end{aligned}

By comparing numerator of the above expression we get

\begin{aligned}-2u &= a\alpha(1-b)+bu\alpha \end{aligned}

Again by comparing the coefficient of $u$ and constant term from both side of the above equation we get

$$\left\{\begin{array}{rl}a\alpha(1-b)&=0\\ b\alpha&=-2 \end{array}\right.$$

From last equation we get $u=-\dfrac{2}{\alpha}\implies \alpha=-\dfrac{2}{b^2}$.

Hence the correction option is (C).

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