Q246 / 270 · NDA 2022 (Phase 2) · Mathematics

What is $\int_{-\pi / 2}^{\pi / 2}\left(e^{\cos x} \sin x+e^{\sin x} \cos x\right) d x$ equal to?

  • Correct answer
    A
    $\frac{e^2-1}{e}$
  • B
    $\frac{e^2+1}{e}$
  • C
    $\frac{1-e^2}{e}$
  • D
    $0$

Solution

$\begin{aligned} & I=\int_{-\pi / 2}^{\pi / 2}\left(e^{\cos x} \sin x+e^{\sin x} \cos x\right) d x \\ & =\int_{-\pi / 2}^{\pi / 2} e^{\cos x} \cdot \sin x d x \pm \int_{-\pi / 2}^{\pi / 2} e^{\sin x} \cdot \cos x d x \end{aligned}$ $\because \quad e^{\cos x} \cdot \sin x$ is odd function. $\therefore \int_{-\pi / 2}^{\pi / 2} e^{\cos x} \cdot \sin x d x=0$ Let $\sin x=t, \cos x d x=d t$ when $x=\frac{-\pi}{2}, t=-1$, when $x=\frac{\pi}{2}, t=1$ $\therefore \quad \mathrm{I}=\int_{-1}^1 e^t d t=e-e^{-1}=\frac{e^2-1}{e}$

Hence, the correct answer is option: 'A'.

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