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. $(1,3)$
• C. $(-1,3)$
• D. $(3,1)$

Answer 1

The correct option is B $(1,3)$

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