Mathematics

$$\displaystyle \int _{ -\pi /2 }^{ \pi /2 }{ \dfrac { \cos { x }  }{ 1+{ e }^{ x } }  } dx$$


SOLUTION
$$\displaystyle I=\, ^{\frac{\pi }{2}} _{\frac{-\pi }{2}} \int \frac{\cos x}{1+e^{x}}dx$$      ......(i)

$$\displaystyle \therefore I=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \frac{\cos (\frac{\pi }{2}-\frac{\pi }{2}-x)}{1+e(\frac{\pi }{2}-\frac{\pi }{2}-x)}dx$$

$$\displaystyle=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \frac{\cos x}{1+ e^{-x}}dx=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \frac{e^{x}\cos x}{1+e^{x}}dx$$    ....(ii)

Adding equation (i) and (ii), we get
$$\displaystyle \Rightarrow 2I=\, _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\int \cos x dx = [\sin x]_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$$
$$= \displaystyle \sin\frac{\pi }{2}- \sin \left(-\frac{\pi }{2}\right)$$
$$1+1$$
$$=2$$
$$\Rightarrow 2I=2$$
$$\Rightarrow I=1$$
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Subjective Medium Published on 17th 09, 2020
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