Question

If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$ where $c$ is a constant of integration, then the ordered pair $(a,b)$ is equal to

• A. (-1,3)
• B. (3,1)
• C. (1,3)
• D. (1,-3)

Answer 1

Answer: (C)

$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx$
$=\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x{)}^2}}dx$
Let $\sin x+\cos x=t$
$\int \frac{dt}{\sqrt{9-t^2}}={\sin }^{-1}\frac{t}{3}+c$
$={\sin }^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c$
So $a=1,b=3$