Question

If $\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx={\sin }^{-1}\left(\frac{\sin x+\cos x}{a}\right)+c$, then $a$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}\cdot dx$

rearranging the denominator, we get

$\sqrt{9-(1+\sin 2x)}$

$\sqrt{9-(\sin x+\cos x{)}^2}$

now putting the rearranged denominator in the integration, we have

$\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x{)}^2}}dx$

Let $\sin x+\cos x=p$

$(\cos x-\sin x)dx=dp$

$=\int \frac{dp}{\sqrt{9-p^2}}={\sin }^{-1}{(}^p{/}_3)+c\cdot$

$p=\sin x+\cos x\cdot$

$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx={\sin }^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c$

Hence, option 'B' is correct.