Question

Evaluate :

${\int }_0^{\pi /2}\sqrt{1-\cos 2x}dx$

• A. $\sqrt{2}$
• B. $\frac{1}{2}$
• C. $\sqrt{5}$
• D. None of these

Answer 1

Answer: (A)

$\sqrt{2}$

$I={\int }_0^{\pi /2}\sqrt{1-\cos 2x}dx$

$I={\int }_0^{\pi /2}\sqrt{2{\sin }^{2}x}dx$

$I=\sqrt{2}{\int }_0^{\pi /2}|\sin x|dx=\sqrt{2}{\int }_0^{\pi /2}\sin xdx$

$I=\sqrt{2}[-\cos x{]}_0^{\pi /2}$

$I=\sqrt{2}\left[\right(-\cos \frac{\pi }{2})-(-\cos 0\left)\right]=\sqrt{2}(0+1)=\sqrt{2}$