Question

Evaluate: ${\int }_0^{2\pi }{e}^x\cdot \sin (\frac{\pi }{4}+\frac{x}{2})dx$.

Answer 1

$\Rightarrow \int e^x\sin (\pi /4+x/2)$

let $e^x$ be function & $\sin (\pi /4+x/2)$

be function

Now $y=e^x\int \sin (\pi /9+x/2)-\int \sin (\pi /9+x/2)\times \frac{d}{dx}\left(e^x\right)$

$y=-\frac{e^x}{2}\cos (\pi /9+x/2)+\frac{1}{2}\int \cos (\pi /4+x/2)\times e^x$

$y=\frac{-e^x}{2}\cos (\pi /4+\frac{x}{2})+\frac{1}{2}\times \frac{1}{2}{e}^x\sin (\pi /4+x/2)-\frac{1}{2}\times \frac{1}{2}\int e^x\sin (\pi /4+\frac{x}{2})$

$y=\frac{-e^x}{2}\cos (\pi /4+\pi /2)+\frac{1}{4}{e}^x\sin (\frac{\pi }{4}+\frac{\pi }{2})-\frac{1}{4}\underset{y}{\underset{}{\int e^x\sin (\pi /4+\pi /2)}}$

$\frac{8y}{4}=\frac{-e^x}{2}\cos (\pi /4+\pi /2)+\frac{e^x\sin (\frac{\pi }{4}+\frac{\pi }{2})}{4}$

$y=\frac{-2e^x}{5}\cos (\pi /4+\pi /2)+\frac{e^x\sin (\frac{\pi }{4}+\frac{\pi }{2})}{5}$

${\int }_0^{2\pi }y={(-\frac{2e^x}{5}\cos (\pi /4+\pi /2)+\frac{e^x\sin (\frac{\pi }{4}+\frac{\pi }{2})}{5})}^{2\pi }$

$=-\frac{2e^{2\pi }}{5}\cos \pi /4+\frac{e^{2\pi }(-\sin \pi /4)}{5}$

$=-\frac{3e^{2\pi }}{5\sqrt{2}}$