Question

If \[\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} \, dx = \frac{1}{12} \tan^{-1}(3 \tan x) + \text{constant},\]then the maximum value of $a \sin x + b \cos x$ is:

• A. $\sqrt{40}$
• B. $\sqrt{39}$
• C. $\sqrt{42}$
• D. $\sqrt{41}$

Answer 1

The Correct Option is A

\[\int \frac{\sec^2 x \, dx}{a^2 \tan^2 x + b^2}\]
Substitute \( \tan x = t \), so \( \sec^2 x \, dx = dt \).
\[= \int \frac{dt}{a^2 t^2 + b^2}\]
\[= \frac{1}{a^2} \int \frac{dt}{t^2 + \left( \frac{b}{a} \right)^2}\]
\[= \frac{1}{ab} \tan^{-1} \left( \frac{ta}{b} \right) + c\]
\[= \frac{1}{ab} \tan^{-1} \left( \frac{a}{b} \tan x \right) + c\]
From comparison, \( \frac{a}{b} = 3 \).
Given \(ab = 12\).
Solving yields \(a = 6\) and \(b = 2\).
Maximum Value:
The maximum value of \( 6 \sin x + 2 \cos x \) is \( \sqrt{40} \).