Question

\[\int_0^\pi e^{2x} \cdot \sin\left( \frac{\pi}{4} + x \right) dx\]

Answer 1

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