If $\int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}-dx=\frac{1}{12}{\tan }^{-1}(3\tan x)+C\text{, }$then the maximum value of $a \sin x + b \cos x,$ is

Solution:

We have,

$\begin{array}{l} I=\int \frac{1}{a^2{\sin }^{2}x+b^2{\cos }^{2}x}dx\\ \Rightarrow I=\int \frac{{\sec }^{2}x}{b^2+a^2{\tan }^{2}x}dx\\ \Rightarrow I=\frac{1}{a}\int \frac{1}{b^2+(a\tan x{)}^2}d(a\tan x)\\ =\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan x\right)+C\end{array}$

$\therefore ab=12$ and $\frac{a}{b}=3\Rightarrow a^2=36\Rightarrow a=\pm 6$

$\therefore ab=12\Rightarrow b=\pm 2$

Thus, we have

       $a\sin x+b\cos x=\pm (6\sin x+2\cos x)$

We know that

                  $-\sqrt{a^2+b^2}\le a\sin x+b\cos x\le \sqrt{a^2+b^2}$ for all $x$

$\therefore -\sqrt{40}\le \pm 6\sin x\pm 2\cos x\le \sqrt{40}$ for all $x$.