Question

Evaluate: \[ \int \frac{1}{1+3\sin^2 x+8\cos^2 x}\,dx \]

• A. \(\dfrac{1}{6}\tan^{-1}(2\tan x)+C\)
• B. \(\tan^{-1}(2\tan x)+C\)
• C. \(\dfrac{1}{6}\tan^{-1}\!\left(\dfrac{2\tan x}{3}\right)+C\)
• D. None of these

Hint:

Convert trigonometric expressions using identities before substitution.

Answer 1

The Correct Option is A

Step 1: Write denominator using \(\sin^2x+\cos^2x=1\): \[ 1+3\sin^2x+8\cos^2x=1+3(1-\cos^2x)+8\cos^2x=4+5\cos^2x \]
Step 2: \[ \int \frac{dx}{4+5\cos^2x} \] Put \(\tan x=t\), then simplify.
Step 3: Final integration gives \[ \frac{1}{6}\tan^{-1}(2\tan x)+C \]