Question

$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx$ is equal to

• A. ${\sin }^{-1}(\sin x+\cos x)+c$
• B. ${\sin }^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c$
• C. ${\cos }^{-1}(\sin x+\cos x)+c$
• D. None of these

Answer 1

The correct option is B ${\sin }^{-1}\left[\frac{1}{3}(\sin x+\cos x)\right]+c$

Let $I=\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx$

Substitute $\cos x-\sin x=z\Rightarrow (-\sin x+\cos x)dx=dz$

$\therefore I=\int \frac{dz}{\sqrt{8-(z^2-1)}}$

$=\int \frac{dz}{\sqrt{9-z^2}}$

$={\sin }^{-1}\frac{z}{3}+c$ ............. Using $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

$={\sin }^{-1}\left(\frac{\cos x+\sin x}{3}\right)+c$