Question

$\underset{0}{\overset{2\mathrm{\pi }}{\int }}{e}^{x/2}\sin \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{2\mathrm{\pi }}{e}^{\frac{x}{2}}\sin \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)\mathit{d}x.\mathrm{Then}, \\ \mathrm{Integrating}\mathrm{by}\mathrm{parts} \\ I={\left[-2e^{\frac{x}{2}}\cos \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)\right]}_0^{2\mathrm{\pi }}-{\int }_0^{2\mathrm{\pi }}-\frac{2}{2}{e}^{\frac{x}{2}}\cos \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)dx \\ \mathrm{Again},\mathrm{integrating}\mathrm{second}\mathrm{term}\mathrm{by}\mathrm{parts} \\ \Rightarrow I={\left[-2e^{\frac{x}{2}}\cos \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)\right]}_0^{2\mathrm{\pi }}+\left\{{\left[2e^{\frac{x}{2}}\sin \mathit{}\left(\frac{x}{2}\mathit{+}\frac{\mathrm{\pi }}{4}\right)\right]}_{\mathit{0}}^{2\mathrm{\pi }}\mathit{-}{\int }_{\mathit{0}}^{2\mathrm{\pi }}\frac{2}{2}\mathit{}{e}^{\frac{x}{2}}\sin \mathit{}\left(\frac{x}{2}\mathit{+}\frac{\mathrm{\pi }}{4}\right)\mathit{}\mathit{d}x\right\} \\ \Rightarrow I={\left[-2e^{\frac{x}{2}}\cos \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)\right]}_0^{2\mathrm{\pi }}+{\left[2e^{\frac{x}{2}}\sin \left(\frac{x}{2}+\frac{\mathrm{\pi }}{4}\right)\right]}_0^{2\mathrm{\pi }}-I \\ \Rightarrow 2I=\frac{2}{\sqrt{2}}{e}^{\mathrm{\pi }}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{2}}{e}^{\mathrm{\pi }}-\frac{2}{\sqrt{2}}=0 \\ \Rightarrow I=0 \end{gathered}$$