(a)OR(b)

Evaluate : $\displaystyle \int_{0}^{2 \pi} \frac{1}{1+\mathrm{e}^{\sin \mathrm{x}}} \mathrm{~d} \mathrm{x}$

Let $\displaystyle \mathrm{I}=\int_{0}^{2 \pi} \frac{1}{1+\mathrm{e}^{\sin \mathrm{x}}} \mathrm{d} \mathrm{x}=\int_{0}^{2 \pi} \frac{1}{1+\mathrm{e}^{\sin (2 \pi-\mathrm{x})}} \mathrm{d} \mathrm{x}$=\int_{0}^{2 \pi} \frac{1}{1+\mathrm{e}^{-\sin \mathrm{x}}} \mathrm{dx}=\int_{0}^{2 \pi} \frac{\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{e}^{\sin \mathrm{x}}+1} \mathrm{dx} \Rightarrow 2 \mathrm{I}=\int_{0}^{2 \pi} \frac{\mathrm{e}^{\sin \mathrm{x}}+1}{\mathrm{e}^{\sin \mathrm{x}}+1} \mathrm{dx}=\int_{0}^{2 \pi} 1 \cdot \mathrm{dx}=2 \pi \Rightarrow \quad \mathbf{I}=\pi \begin{array}{r} \mathrm{I}=\int \frac{\mathrm{x}^{4}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+1\right)} \mathrm{d} \mathrm{x}=\int\left[\mathrm{x}+1+\frac{1}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+1\right)}\right] \mathrm{d} \mathrm{x} =\frac{\mathrm{x}^{2}}{2}+\mathrm{x}+\int\left[\frac{1}{2(\mathrm{x}-1)}-\frac{1}{2} \frac{(\mathrm{x}+1)}{\left(\mathrm{x}^{2}+1\right)}\right] \mathrm{d} \mathrm{x} \end{array}(Using partial fractions)=\frac{\mathrm{x}^{2}}{2}+\mathrm{x}+\frac{1}{2} \log |\mathrm{x}-1|-\frac{1}{4} \log \left|\mathrm{x}^{2}+1\right|-\frac{1}{2} \tan ^{-1} \mathrm{x}+C $$