Question

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

Answer 1

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

$I={\int }_0^{2\pi }{e}^x\sin (\frac{\pi }{4}+\frac{x}{2})dx$

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

$I=\sin (\frac{\pi }{4}+\frac{x}{2})e^x-\frac{1}{2}\left[cos\right(\frac{\pi }{4}+\frac{x}{2})e^x+\frac{1}{2}\int \sin (\frac{\pi }{4}+\frac{x}{2}\left)e^{x}dx\right]$

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

$I=\sin (\frac{\pi }{4}+\frac{x}{2})e^x-\frac{1}{2}\cos (\frac{\pi }{4}+\frac{x}{2})e^x-\frac{I}{4}$

$I+\frac{I}{4}=\sin (\frac{\pi }{4}+\frac{x}{2})e^x{|}_0^{2\pi }-\frac{1}{2}cos(\frac{\pi }{4}+\frac{x}{2})e^x{|}_0^{2\pi }$

$\frac{5I}{4}=\sin (\frac{\pi }{4}+\pi ).e^{2\pi }-\sin \left(\frac{\pi }{4}\right).e^0-\frac{1}{2}\left[cos\right(\frac{\pi }{4}+\pi )e^{2\pi }-\cos (\frac{\pi }{4}).e^0]$

$\frac{5I}{4}=\frac{-1}{\sqrt{2}}{e}^{2\pi }-\frac{1}{\sqrt{2}}.1+\frac{1}{2}.\frac{1}{\sqrt{2}}{e}^{2\pi }+\frac{1}{\sqrt{2}}.\frac{1}{2}$

$\frac{5I}{4}=e^{2\pi }(\frac{-1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}})-\frac{1}{2\sqrt{2}}$

$\frac{5I}{4}=\frac{-1}{2\sqrt{2}}{e}^{2\pi }-\frac{1}{2\sqrt{2}}=\frac{-1}{2\sqrt{2}}(e^{2\pi }+1)$

$I=\frac{4}{5}\times \frac{-1}{2\sqrt{2}}(e^{2\pi }+1)$

$I=\frac{-\sqrt{2}}{5}(e^{2\pi }+1)$