Question

$\int _ { - \pi / 2 } ^ { \pi / 2 } \frac { \sin x } { 1 + \cos ^ { 2 } x } e ^ { - \cos ^ { 2 } x } d x$ is equal to

• A. $2 e ^ { - 1 }$
• B. 1
• C. 0
• D. None of these

Answer 1

Answer: (C)

0

Answer 2

$I = \int _ { - \pi / 2 } ^ { \pi / 2 } \frac { \sin x } { 1 + \cos ^ { 2 } x } e ^ { - \cos ^ { 2 } x } d x$

$\because \frac { \sin x } { 1 + \cos ^ { 2 } x } e ^ { - \cos ^ { 2 } x }$is an odd function,

∴$I = 0$.