If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} \mathrm{~d} x=\operatorname{asin}^{-1}\left(\frac{\sin…
If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} \mathrm{~d} x=\operatorname{asin}^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$
Where c is a constant of integration, then the ordered pair $(\mathrm{a}, \mathrm{b})$ is equal to
$(1,3)$
$(3,1)$
$(-1,3)$
$(-3,1)$
Solution
$\begin{aligned}
\text {Let } \mathrm{I} & =\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x \\
& =\int \frac{\cos x-\sin x}{\sqrt{9-(1+\sin 2 x)}} d x \\
& =\int \frac{\cos x-\sin x}{\sqrt{3^2-(\cos x+\sin x)^2}} \mathrm{~d} x
\end{aligned}$ Put $\cos x+\sin x=\mathrm{t}$
$\Rightarrow(-\sin x+\cos x) \mathrm{d} x=\mathrm{dt}$
$\begin{aligned}
\therefore \quad I & =\int \frac{d t}{\sqrt{3^2-t^2}} d t \\
& =\sin ^{-1}\left(\frac{t}{3}\right)+c \\
& =\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c \\
\therefore \quad & (a, b)=(1,3)
\end{aligned}$