Question

\[\int\limits_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]

Answer 1

\[Let\ I = \int_0^{2\pi} e^x \cos\left( \frac{\pi}{4} + \frac{x}{2} \right) d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ 2 e^x \sin \left( \frac{\pi}{4} + \frac{x}{2} \right) \right]_0^{2\pi} - \int_0^{2\pi} 2 e^x \sin \left( \frac{\pi}{4} + \frac{x}{2} \right) dx\]
\[\text{Integrating second term by parts}\]
\[I = \left[ 2 e^x \sin \left( \frac{\pi}{4} + \frac{x}{2} \right) \right]_0^{2\pi} + \left\{ \left[ 4 e^x \cos \left( \frac{\pi}{4} + \frac{x}{2} \right) \right]_0^{2\pi} + \int_0^{2\pi} - 4 e^x \cos \left( \frac{\pi}{4} + \frac{x}{2} \right) d x \right\}\]
\[ \Rightarrow I = \left[ 2 e^x \sin \left( \frac{\pi}{4} + \frac{x}{2} \right) \right]_0^{2\pi} + \left[ 4 e^x \cos \left( \frac{\pi}{4} + \frac{x}{2} \right) \right]_0^{2\pi} - 4I\]
\[ \Rightarrow 5I = - 2 e^{2\pi} \frac{1}{\sqrt{2}} - 2 \frac{1}{\sqrt{2}} - 4 e^{2\pi} \frac{1}{\sqrt{2}} - 4 \frac{1}{\sqrt{2}}\]
\[ \Rightarrow 5I = - 3\sqrt{2} e^{2\pi} - 3\sqrt{2}\]
\[ \Rightarrow I = - \frac{3\sqrt{2}}{5}\left( e^{2\pi} + 1 \right)\]