Question

Evaluate the integral: \[ \int \frac{dx}{9\cos^2 2x + 16\sin^2 2x} \]

• A. \( \frac{1}{25} \tan^{-1} \left( \frac{3}{4} \sec^2 2x \right) + c \)
• B. \( \frac{1}{25} \tan^{-1} \left( \frac{4}{3} \sec^2 2x \right) + c \)
• C. \( \frac{1}{24} \tan^{-1} \left( \frac{3}{4} \tan 2x \right) + c \)
• D. \( \frac{1}{24} \tan^{-1} \left( \frac{4}{3} \tan 2x \right) + c \)

Hint:

Express denominator in standard quadratic form for easier integration. - Use substitution \( t = \tan x \) for rational trigonometric integrals.

Answer 1

The Correct Option is D

Step 1: Express the denominator
Rewriting, \[ 9\cos^2 2x + 16\sin^2 2x = 9 + 7\sin^2 2x. \] Substituting \( t = \tan 2x \), we get: \[ dt = 2\sec^2 2x dx. \] Step 2: Integrate using substitution
\[ I = \int \frac{dx}{9 + 7\tan^2 2x}. \] Using standard integration formula: \[ I = \frac{1}{24} \tan^{-1} \left( \frac{4}{3} \tan 2x \right) + c. \] Thus, the correct answer is \( \boxed{\frac{1}{24} \tan^{-1} \left( \frac{4}{3} \tan 2x \right) + c} \).