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:

Use substitution $t = a + b\cos x$ for such integrals.

Answer 1

The Correct Option is D

Concept:
Use substitution method with trigonometric identity. Step 1: Rewrite numerator. \[ \sin 2x = 2\sin x \cos x \]
Step 2: Substitute. Let: \[ t = a + b\cos x \Rightarrow dt = -b\sin x dx \] \[ \sin x dx = \frac{-dt}{b} \]
Step 3: Transform integral. \[ I = \int \frac{2\sin x \cos x}{(a+b\cos x)^2} dx = \int \frac{2\cos x}{t^2} \cdot \frac{-dt}{b} \] \[ = -\frac{2}{b} \int \frac{\cos x}{t^2} dt \] \[ \cos x = \frac{t-a}{b} \] \[ I = -\frac{2}{b^2} \int \frac{t-a}{t^2} dt \]
Step 4: Split integral. \[ = -\frac{2}{b^2} \int \left(\frac{1}{t} - \frac{a}{t^2}\right) dt \] \[ = -\frac{2}{b^2} \left[\log t + \frac{a}{t}\right] \]
Step 5: Back substitute. \[ = -\frac{2}{b^2} \left[\log(a + b\cos x) - \frac{a}{a+b\cos x}\right] + c \]
Step 6: Conclusion. Correct option is (D).