Step 1: Understanding the Concept:
The integrand contains a double angle $\sin 2x$ in the numerator and a linear term in $\cos x$ squared in the denominator. To make substitution feasible, we must first express the numerator entirely in terms of single angles $x$. Then, substituting the entire linear denominator block makes the integration straightforward.
Step 2: Key Formula or Approach:
1. Double angle identity: $\sin 2x = 2 \sin x \cos x$.
2. Substitution method: Let $u = a + b\cos x$. Find $du$ and express the remaining $\cos x$ in terms of $u$.
3. Power rule for integration: $\int u^n du = \frac{u^{n+1}}{n+1}$ (for $n \neq -1$) and $\int \frac{1}{u} du = \log|u|$.
Step 3: Detailed Explanation:
Let the integral be $I = \int \frac{\sin 2x}{(a+b\cos x)^2} dx$.
Use the identity $\sin 2x = 2 \sin x \cos x$:
\[ I = \int \frac{2 \sin x \cos x}{(a+b\cos x)^2} dx \]
Now, apply substitution. Let $u = a + b\cos x$.
Then, differentiate with respect to $x$: $du = -b \sin x \,dx \implies \sin x \,dx = -\frac{du}{b}$.
We also need an expression for the lone $\cos x$ in the numerator. From our substitution, we get $\cos x = \frac{u - a}{b}$.
Substitute everything into the integral:
\[ I = \int \frac{2 \left(\frac{u - a}{b}\right)}{u^2} \left( -\frac{du}{b} \right) \]
Extract the constants to the front:
\[ I = -\frac{2}{b^2} \int \frac{u - a}{u^2} du \]
Split the fraction in the integrand:
\[ I = -\frac{2}{b^2} \int \left( \frac{u}{u^2} - \frac{a}{u^2} \right) du \]
\[ I = -\frac{2}{b^2} \int \left( \frac{1}{u} - a u^{-2} \right) du \]
Now, integrate each term with respect to $u$:
\[ I = -\frac{2}{b^2} \left[ \log|u| - a \left( \frac{u^{-1}}{-1} \right) \right] + c \]
\[ I = -\frac{2}{b^2} \left[ \log|u| + \frac{a}{u} \right] + c \]
Finally, substitute back $u = a + b\cos x$:
\[ I = -\frac{2}{b^2} \left[ \log(a + b\cos x) + \frac{a}{a+b\cos x} \right] + c \]
(Assuming the argument of the logarithm is positive in the given domain, absolute values can be dropped to match the options).
Step 4: Final Answer:
The result is $\frac{-2}{b^2} \left[\log(a + b \cos x) + \frac{a}{a+b\cos x}\right] + c$.