Question

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

• A. $\pi$
• B. $0$
• C. $-\pi /2$
• D. $2\pi$

Answer 1

Answer: (B)

$0$

Substitute $x/2=\theta$, so that
$I=2{\int }_0^{\pi }{e}^{\theta }\sin (\theta +\frac{\pi }{4})d\theta$
$=\sqrt{2}{\int }_0^{\pi }{e}^{\theta }(\sin \theta +\cos \theta )d\theta$
$={\sqrt{2}{e}^{\theta }\sin \theta ]}_0^{\pi }=0$
$[\because \int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx=e^{x}f(x\left)\right]$