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If $$\int \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx = \frac{1}{12} \tan^{-1}(3 \tan x) +$$ constant, then the maximum value of $$a \sin x + b \cos x$$, is :
$$\int \frac{dx}{a²sin²x+b²cos²x} = \frac{1}{ab}tan⁻¹(\frac{a}{b}tanx) + C$$.
Comparing: 1/(ab) = 1/12 and a/b = 3. So ab = 12, a = 3b. 3b² = 12, b² = 4, b = 2, a = 6.
Max of a sin x + b cos x = √(a²+b²) = √(36+4) = √40.
The correct answer is Option (1): √40.
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