MCQ
$\int_{-\pi / 2}^{\pi / 2} \frac{\sin x}{1+\cos ^2 x} e^{-\cos ^2 x} d x$ is equal to
  • A
    $2 e^{-1}$
  • B
    1
  • $0$
  • D
    -1

Answer

Correct option: C.
$0$
(C)
Let $f (x)=\frac{\sin x}{1+\cos ^2 x} e ^{-\cos ^2 x}$
$\begin{array}{ll}\therefore & f (-x)=-\frac{\sin x}{1+\cos ^2 x} e ^{-\cos ^2 x}=- f (x) \\ \therefore & f (x) \text { is an odd function. } \\ \therefore & \int_{-\pi / 2}^{\pi / 2} f (x) d x=0\end{array}$

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