Question

$\int _ { - \pi / 2 } ^ { \pi / 2 } \sqrt { \frac { 1 } { 2 } ( 1 - \cos 2 x ) } d x =$

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

Answer 1

Answer: (B)

2

Answer 2

$\int _ { - \pi / 2 } ^ { \pi / 2 } \sqrt { \frac { 1 } { 2 } ( 1 - \cos 2 x ) } d x = 2 \int _ { 0 } ^ { \pi / 2 } | \sin x | d x$

= $2 [ - \cos x ] _ { 0 } ^ { \pi / 2 } = 2 \left[ - \cos \left( \frac { \pi } { 2 } \right) + \cos 0 \right] = 2$.