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 :

If $$\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C$$, where $$C$$ is the integration constant, then $$A B$$ is equal to

For $$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, if $$y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x$$, and $$\lim _\limits{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$$ then $$y\left(\frac{\pi}{4}\right)$$ is equal to

$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$