Question

If $f ( x )$ is an odd function of $x$ then $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } f ( \cos x ) d x$ is equal to

• A. 0
• B. $\int _ { 0 } ^ { \frac { \pi } { 2 } } f ( \cos x ) d x$
• C. $2 \int _ { 0 } ^ { \frac { \pi } { 2 } } f ( \sin x ) d x$
• D. $\int _ { 0 } ^ { \pi } f ( \cos x ) d x$

Answer 1

Answer: (C)

$2 \int _ { 0 } ^ { \frac { \pi } { 2 } } f ( \sin x ) d x$

Answer 2

$f ( \cos x )$ is an even function.

$\because f ( \cos ( - x ) ) = f ( \cos x )$

$\therefore \int _ { - \pi / 2 } ^ { \pi / 2 } f ( \cos x ) d x = 2 \int _ { 0 } ^ { \pi / 2 } f ( \cos x ) d x$ $= 2 \int _ { 0 } ^ { \pi / 2 } f ( \sin x ) d x$.