Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade None
Question:
$\int \frac{\sin 2x}{8\sin^2 x + 17\cos^2 x}dx$ is equal to:
-\frac{1}{9}\log_e(8\sin^2 x + 17\cos^2 x) + c
\log_e \frac{1}{\sqrt[8]{8 + 9\cos^2 x}} + c
\log_e \frac{1}{\sqrt[8]{17 - 9\sin^2 x}} + c
None of these
Step-by-Step Solution
Key Concept: Recognize the denominator's derivative is proportional to $\sin 2x$ in the numerator, making this a direct substitution integral.
Let $u = 8\sin^2 x + 17\cos^2 x$. Then $\frac{du}{dx} = 16\sin x \cos x - 34\cos x \sin x = -18\sin x \cos x = -9\sin 2x$. Therefore $\sin 2x \, dx = -\frac{1}{9}du$. Substituting: $\int \frac{\sin 2x}{8\sin^2 x + 17\cos^2 x}dx = \int \frac{1}{u} \cdot (-\frac{1}{9})du = -\frac{1}{9}\ln|u| + c = -\frac{1}{9}\ln(8\sin^2 x + 17\cos^2 x) + c$. This equals options 2 and 3 since $8\sin^2 x + 17\cos^2 x = 8 + 9\cos^2 x = 17 - 9\sin^2 x$ (using $\sin^2 x + \cos^2 x = 1$).
Correct Answer: 1,2,3