Question

The value of ${\int }_0^{\pi }\frac{1}{1+e^{\cos x}}\mathrm{dx}$ is

• A. $\pi$
• B. \N
• C. $\frac{\pi }{2}$
• D. $-\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{2}$

${\int }_0^{\pi }\frac{1}{1+e^{\cos x}}\mathrm{dx}={\int }_0^{\pi /2}\left(\frac{1}{1+e^{\cos x}}+\frac{1}{1+e^{\cos \left(\pi -x\right)}}\right\}\mathrm{dx}\left(\mathrm{As}{\int }_0^{2a}f\left(x\right)\mathrm{dx}={\int }_0^{a}f\left(x\right)\mathrm{dx}+{\int }_0^{a}f(a-x)\mathrm{dx}\right]={\int }_0^{\pi /2}\left(\frac{1}{1+e^{\cos x}}+\frac{1}{1+e^{-\cos x}}\right\}\mathrm{dx}={\int }_0^{\pi /2}\left(\frac{1}{1+e^{\cos x}}+\frac{e^{\cos x}}{1+e^{\cos x}}\right\}\mathrm{dx}={\int }_0^{\pi /2}\left(\frac{1+e^{\cos x}}{1+e^{\cos x}}\right\}\mathrm{dx}=={\int }_0^{\pi /2}1\cdot \mathrm{dx}={\left(x\right]}_0^{\pi /2}=\frac{\pi }{2}-0=\frac{\pi }{2}$