Question

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

• A. $log(\cos x+\sec x+\sqrt{sin2x})+c$
• B. $log(\cos x+\sin x+\sqrt{sin2x})+c$
• C. $log(\cos x+\sec x+\sqrt{cos2x})+c$
• D. $-log(\cos x+\sin x+\sqrt{sin2x})+c$

Answer 1

Answer: (D)

$-log(\cos x+\sin x+\sqrt{sin2x})+c$

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

$(\cos x+\sin x{)}^2={\cos }^{2}x+{\sin }^{2}x+2\cos x\sin x$

$=1+\sin 2x$

$\therefore \sin 2x=(\cos x+\sin x{)}^2-1$

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

Let, $t=\cos x+\sin x$

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

From $\left(1\right)$

$\int \frac{\sin x-\cos x}{\sqrt{\sin 2x}}dx=-\frac{1}{\sqrt{t^2-1}}dt$

$=-log|t+\sqrt{t^2-1}|+c$

$=-log|\sin x+\cos x+\sqrt{(\sin x+\cos x{)}^2-1}|+c$

$\therefore \int \frac{\sin x-\cos x}{\sqrt{\sin 2x}}dx=-log|\sin x+\cos x+\sqrt{\sin 2x}|+c$.