Mathematics

$$\text { Evaluate: } \displaystyle \int_{-\pi / 2}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} \mathrm{d} \mathrm{x}$$


SOLUTION
Let 
$$\begin{array}{l|c}I=\displaystyle \int_{-\pi / 2}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} d x \\=\displaystyle \int_{-\pi / 2}^{0} \dfrac{\cos x}{1+e^{x}} d x+\displaystyle \int_{0}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} d x \\=\displaystyle \int_{\pi / 2}^{0} \dfrac{\cos t}{1+e^{-t}}(-d t)+ \displaystyle \int_{0}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} d x \hspace{3cm}[\text{Substituting  }\ x=-t]\\=\displaystyle \int _6^{\pi / 2} \dfrac{\cos t}{1+\dfrac{1}{e^{t}}} d t+\displaystyle\int_{0}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} d x=\displaystyle \int_{0}^{\pi / 2} \dfrac{e^{t} \cdot \cos t}{1+e^{t}} d t+\displaystyle \int_{0}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} d x \\=\displaystyle \int_{0}^{\pi / 2} \dfrac{e^{x} \cdot \cos x}{1+e^{x}} d x+\displaystyle \int_{0}^{\pi / 2} \dfrac{\cos x}{1+e^{x}} d x \\=\displaystyle \int_{0}^{\pi / 2} \dfrac{\left(e^{x}+1\right) \cdot \cos x}{1+e^{x}} d x=\int_{0}^{\pi / 2} \cos x d x \\=\left[\sin\ x\right]_0^{\pi/2}\\=\sin \pi / 2-\sin 0 \\=1\end{array}$$
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Subjective Medium Published on 17th 09, 2020
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