Question

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

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

Answer 1

The correct option is D $-\frac{3\sqrt{2}}{5}(e^{2\pi }+1)$

We have, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$ .............. Integration by parts

Let $I={\int }_0^{2\pi }{e}^x\cos (\frac{\pi }{4}+\frac{x}{2})dx$
$={\left[e^x\cos (\frac{\pi }{4}+\frac{x}{2})\right]}_0^{2\pi }+\frac{1}{2}{\int }_0^{2\pi }{e}^x\sin (\frac{\pi }{4}+\frac{x}{2})dx$ .......... Using Integration by parts
$={\left[e^x\cos (\frac{\pi }{4}+\frac{x}{2})\right]}_0^{2\pi }+\frac{1}{2}{\left[e^x\sin (\frac{\pi }{4}+\frac{x}{2})\right]}_0^{2\pi }-\frac{1}{4}{\int }_0^{2\pi }{e}^x\cos (\frac{\pi }{4}+\frac{x}{2})dx$ .......... Using Integration by parts

$\Rightarrow \frac{5}{4}I=({-e}^{2\pi }\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})+\frac{1}{2}(-e^{2\pi }\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}})$
$\Rightarrow I=-\frac{3\sqrt{2}}{5}(e^{2\pi }+1)$