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 :
Step-by-Step Solution
Key Concept: Divide numerator and denominator by cos^2 x to convert the integral into the form integral of sec^2 x / (a^2 tan^2 x + b^2) dx. Use substitution u = tan x and compare with the standard integral formula 1/(ab) tan^-1((a/b) tan x).
The integral is $\int \frac{\sec^2 x}{a^2 \tan^2 x + b^2} dx$. Let $u = \tan x$, then $du = \sec^2 x dx$. The integral becomes $\int \frac{du}{a^2 u^2 + b^2} = \frac{1}{ab} \tan^{-1}(\frac{a}{b} u) + C = \frac{1}{ab} \tan^{-1}(\frac{a}{b} \tan x) + C$. Comparing with $\frac{1}{12} \tan^{-1}(3 \tan x)$, we get $ab = 12$ and $a/b = 3$. Thus $a^2 = 3b^2$, so $3b^2 = 12 \implies b^2 = 4 \implies b = 2$ and $a^2 = 3(4) = 12 \implies a = \sqrt{12} = 2\sqrt{3}$. The maximum value of $a \sin x + b \cos x$ is $\sqrt{a^2 + b^2} = \sqrt{12 + 4} = \sqrt{16} = 4$. Wait, checking the options, $\sqrt{40}$ is option 1. Let's re-evaluate. If $a^2=36, b^2=4$, then $ab=12$ and $a/b=3$. Then $a=6, b=2$. Max value is $\sqrt{36+4} = \sqrt{40}$.
Correct Answer: (1)