Question

The value of ${\int }_{-\pi /2}^{\pi /2}\sqrt{\frac{1}{2}(1-\cos 2x)}$ dx is

• A. $0$
• B. $2$
• C. $\frac{1}{2}$
• D. None of these.

Answer 1

Answer: (B)

$2$

$I={\int }_{-\pi /2}^{\pi /2}\sqrt{\frac{1}{2}.2{\sin }^{2}x}dx={\int }_{-\pi /2}^{\pi /2}\left|\sin x\right|dx,$
$=2{\int }_0^{-\pi /2}\left|\sin x\right|dx$
$=2{\int }_0^{-\pi /2}\sin xdx,$ $\because \left|\sin x\right|=\sin x$ on the interval $(0,\pi /2)$
$=2{(-\cos x)}_0^{\pi /2}=2$
Ans: B