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

Answer: (A)

$\sqrt{40}$

Answer 2

$\int \frac{\sec^2 x \, dx}{a^2 \tan^2 x + b^2}$
Let $\tan x = t$, then $\sec^2 x \, dx = dt$.
$= \int \frac{dt}{a^2 t^2 + b^2}$
$= \frac{1}{a^2} \int \frac{dt}{t^2 + \left( \frac{b}{a} \right)^2}$
$= \frac{1}{ab} \tan^{-1} \left( \frac{ta}{b} \right) + c$
$= \frac{1}{ab} \tan^{-1} \left( \frac{a}{b} \tan x \right) + c$
On comparing, $\frac{a}{b} = 3$.
$ab = 12$
$a = 6, \quad b = 2$
Maximum Value:
The maximum value of $6 \sin x + 2 \cos x$ is $\sqrt{40}$.