If $\int \frac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\frac{1}{12} \tan ^{-1}(3 \tan x)+$ constant, then the maximum value of $\mathrm{a} \sin x+\mathrm{b} \cos x$, is :
$\sqrt{40}$
$\sqrt{41}$
$\sqrt{39}$
$\sqrt{42}$
Solution
$\begin{aligned}
& \int \frac{\sec ^2 x d x}{a^2 \tan ^2 x+b^2} \\
& \text { let } \tan x=t \\
& \sec ^2 d x=d t \\
& \int \frac{d t}{a^2 t^2+b^2} \\
& \frac{1}{a^2} \int \frac{d t}{t^2+\left(\frac{b}{a}\right)^2} \\
& \frac{1}{a^2} \frac{1}{\frac{b}{a}} \tan ^{-1}\left(\frac{t}{b} a\right)+c \\
& \frac{1}{a b} \tan ^{-1}\left(\frac{\alpha}{b} \tan x\right)+c
\end{aligned}$
on comparing $\frac{\mathrm{a}}{\mathrm{b}}=3$
$\begin{aligned}
& a b=12 \\
& a=6, b=2
\end{aligned}$
maximum value of
$6 \sin x+2 \cos x \text { is } \sqrt{40}$