Question

Evaluate $\int \sqrt{\sin 2x}dx$

• A. $I=\frac{1}{2}{\text{sin}}^{-1}\sqrt{\sin 2x}+\frac{\sqrt{\sin 2x-{\sin }^{2}2}x}{2}+c$
• B. $I=\frac{1}{2}{\text{sin}}^{-1}\sqrt{\cos 2x}-\frac{\sqrt{\sin 2x-{\sin }^{2}2}x}{2}+c$
• C. $I=\frac{1}{2}{\text{sin}}^{-1}\sqrt{\sin 2x}-\frac{\sqrt{\sin 2x-{\sin }^{2}2}x}{2}+c$
• D. None of these

Answer 1

The correct option is C $I=\frac{1}{2}{\text{sin}}^{-1}\sqrt{\sin 2x}-\frac{\sqrt{\sin 2x-{\sin }^{2}2}x}{2}+c$

Put $\sin 2x=t^2⟹tdt=2\cos 2xdx⟹dx=\frac{tdt}{\sqrt{1-t^2}}$

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

$={\text{sin}}^{-1}t-\frac{t}{2}\sqrt{1-t^2}-\frac{1}{2}{\text{sin}}^{-1}t+c$

$I=\frac{1}{2}{\text{sin}}^{-1}\sqrt{\sin 2x}-\frac{\sqrt{\sin 2x-{\sin }^{2}2}x}{2}+c$