Question:
If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x=a \sin ^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$, where $c$ is a constant of integration, then the ordered pair $(a, b)$ is equal to :
Correct Option: , 2
Solution:
Put $\sin x+\cos x=t$
$\Rightarrow 1+\sin 2 x=t^{2}$
$\Rightarrow(\cos x-\sin x) d x=d t$
$\therefore I=\int \frac{d t}{\sqrt{8-\left(t^{2}-1\right)}}=\int \frac{d t}{\sqrt{9-t^{2}}}$
$=\sin ^{-1}\left(\frac{t}{3}\right)+C=\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c$
$\Rightarrow a=1$ and $b=3$