Question

Solve $\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

• A. $2{\sin }^{-1}\left[\sqrt{2}\sin (x-\frac{\pi }{4})\right]+c$
• B. ${\sin }^{-1}\left[\sqrt{2}\sin (x-\frac{\pi }{4})\right]+c$
• C. ${\sin }^{-1}\left[\sqrt{2}\sin (x+\frac{\pi }{4})\right]+c$
• D. None of these

Answer 1

Answer: (B)

${\sin }^{-1}\left[\sqrt{2}\sin (x-\frac{\pi }{4})\right]+c$

$I=\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

Let $\sin x-\cos x=t$

$(\cos x+\sin x)dx=dt$

$t^2={\sin }^{2}x+{\cos }^{2}x-2\sin xcosx$

$1-\sin 2x$

$\sin 2x=1-t^2$

$=\int \frac{dt}{\sqrt{1-t^2}}$

$={\sin }^{-1}|t|+c$

$={\sin }^{-1}(\sin x-\cos x)+c$

${\sin }^{-1}\left[\sqrt{2}\right(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\left)\right]+c$

${\sin }^{-1}\left[\sqrt{2}\sin \right(x-\frac{\pi }{4}\left)\right]+c$