$$\begin{aligned} & \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+c \\ & I=\int \frac{\sec ^2 x}{b^2+a^2 \tan ^2 x} d x \\ & \tan x=t \\ & \Rightarrow \sec ^2 x d x=d t \\ & I=\int \frac{d t}{b^2+a^2 t^2} \\ & =\frac{1}{b a} \tan ^{-1}\left(\frac{a t}{b}\right)+c \\ & I=\frac{1}{a b} \tan ^{-1}\left(\frac{a}{b} \tan x\right)+c \\ & \Rightarrow a b=12 \text { and } \frac{a}{b}=3\\ & \Rightarrow a^2=36 \text { and } b^2=4\\ & \Rightarrow \text { Maximum value of } a \sin x+b \cos x \text { is } \sqrt{a^2+b^2} \end{aligned}$$

$$\quad=\sqrt{36+4}$$

$$\quad=\sqrt{40}$$