Question

If $I={\int }_0^{2\pi }{e}^x\cos (\frac{x}{2}+\frac{\pi }{4})dx$ then $I$ equals

• A. $\frac{-3}{5}(e^{2\pi }-1)$
• B. $\frac{-3\sqrt{2}}{5}(e^{2\pi }-1)$
• C. $\frac{-3\sqrt{2}}{5}(e^{2\pi }+1)$
• D. $0$

Answer 1

Answer: (C)

$\frac{-3\sqrt{2}}{5}(e^{2\pi }+1)$

Put $\frac{x}{2}+\frac{\pi }{4}=\theta$ so that $x=2\theta -\pi /2$ and
$dx=2d\theta$ Thus,
$I=2{\int }_{\pi /4}^{5\pi /4}{e}^{2\theta -\pi /2}\cos \theta d\theta$
$=2e^{-\pi /2}{I}_1$
where $I_1={\int }_{\pi /4}^{5\pi /4}{e}^{2\theta }\cos \theta d\theta$
$=\frac{1}{2}{e}^{2\theta }\cos \theta {]}_{\pi /4}^{5\pi /4}+\frac{1}{2}{\int }_{\pi /4}^{5\pi /4}{e}^{2\theta }\sin \theta d\theta$
$=\frac{1}{4}{e}^{2\theta }(2\cos \theta +\sin \theta ){]}_{\pi /4}^{5\pi /4}-\frac{1}{4}{I}_1$
$\Rightarrow \frac{5}{4}{I}_1=\frac{1}{4}[e^{5\pi /2}(\frac{-2}{\sqrt{2}}-\frac{1}{\sqrt{2}})-e^{\pi /2}(\frac{2}{\sqrt{2}}+\frac{1}{\sqrt{2}})]$
$\Rightarrow I_1=\frac{-3}{5\sqrt{2}}(e^{5\pi /2}+e^{\pi /2})$
Thus,
$I=\frac{-3\sqrt{2}}{5}(e^{2\pi }+1)$