Question

The value of ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos x}{1+e^x}\mathrm{dx}$ is equal to

Answer 1

$\mathrm{Let}I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\cos x}{1+e^x}\mathrm{dx}\mathrm{Then},I={\int }_0^{\frac{\pi }{2}}\left(\frac{\cos x}{1+e^x}+\frac{\cos \left(-x\right)}{1+e^{-x}}\right\}\mathrm{dx}\left(\mathrm{As}{\int }_{-a}^{a}f\left(x\right)\mathrm{dx}={\int }_0^a\left(f\left(x\right)+f(-x)\right\}\mathrm{dx}\right]\Rightarrow I={\int }_0^{\frac{\pi }{2}}\left(\frac{\cos x}{1+e^x}+\frac{\cos x}{1+e^{-x}}\right\}\mathrm{dx}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\left(\frac{1}{1+e^x}+\frac{e^x}{1+e^x}\right\}\cos x\mathrm{dx}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\left(\frac{1+e^x}{1+e^x}\right\}\cos x\mathrm{dx}\Rightarrow I={\int }_0^{\frac{\pi }{2}}\cos x\mathrm{dx}\Rightarrow I={\left(\sin x\right]}_0^{\frac{\pi }{2}}=\sin \frac{\pi }{2}-\sin 0=1$