Question

$\int \frac{\sin 2x}{(a+b\cos x)^2} dx =$

• A. $\frac{2}{a^2} \left[\log(a + b \cos x) - \frac{a}{a+b\cos x}\right] + c$ where c is the constant of integration.
• B. $\frac{-1}{a^2} \left[\log(a + b \cos x) + \frac{a}{a+b\cos x}\right] + c$, where c is the constant of integration.
• C. $\frac{-2}{b^2} \left[\log(a + b \cos x) + \frac{a}{a+b\cos x}\right] + c$ where c is the constant of integration.
• D. $\frac{-2}{b^2} \left[\log(a + b \cos x) - \frac{a}{a+b\cos x}\right] + c$, where c is the constant of integration.

Hint:

Write sin 2x as 2 sin x cos x and substitute t equal to a plus b cos x.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
\(\sin2x=2\sin x\cos x\). Let \(t=a+b\cos x\), so \(dt=-b\sin x\,dx\) and \(\cos x=\frac{t-a}{b}\).

Step 2: Substitute
\(\int\frac{2\sin x\cos x}{t^2}dx=\int\frac{2(t-a)}{b}\cdot\frac{1}{t^2}\cdot\frac{-dt}{b}=-\frac{2}{b^2}\int\left(\frac1t-\frac{a}{t^2}\right)dt\).

Step 3: Integrate
\(=-\frac{2}{b^2}\left[\log t+\frac{a}{t}\right]+c\).

Step 4: Compare the options
Putting back \(t=a+b\cos x\) gives the third option.

Final Answer:
The integral is \(-\frac{2}{b^2}\left[\log(a+b\cos x)+\frac{a}{a+b\cos x}\right]+c\). \[ \boxed{\text{(C) }\dfrac{-2}{b^{2}}\left[\log(a+b\cos x)+\dfrac{a}{a+b\cos x}\right]+c} \]