Q75 / 180 · KCET 2020 · Mathematics

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ is

  • A
    2
  • B
    0
  • Correct answer
    C
    1
  • D
    $-2$

Solution

We have, $\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{x}} d x$ $=\int_{0}^{\pi / 2}\left(\frac{\cos x}{1+e^{x}}+\frac{\cos (-x)}{\left(1+e^{-x}\right)}\right) d x$ $\left[\because \int_{-a}^{a} f(x) d x=\int_{0}^{a}(f(x)+f(-x)) d x\right]$ $=\int_{0}^{\pi / 2}\left(\frac{\cos x}{1+e^{x}}+\frac{e^{x} \cos x}{1+e^{x}}\right) d x$ $=\int_{0}^{\pi / 2} \cos x\left(\frac{1+e^{x}}{1+e^{x}}\right) d x$ $=\int_{0}^{\pi / 2} \cos x d x=(\sin x)_{0}^{\pi / 2}$ $=\left(\sin \frac{\pi}{2}-\sin 0\right)=1$

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

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