Question

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

• A. $\sqrt{2}{e}^{\frac{x}{2}}\sin \frac{x}{2}+c$
• B. $\sqrt{2}{e}^{\frac{x}{2}}\cos \frac{x}{2}+c$
• C. $-\sqrt{2}{e}^{\frac{x}{2}}\sin \frac{x}{2}+c$
• D. $-\sqrt{2}{e}^{\frac{x}{2}}\cos \frac{x}{2}+c$

Answer 1

The correct option is A $\sqrt{2}{e}^{\frac{x}{2}}\sin \frac{x}{2}+c$

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

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

Put $\frac{x}{2}=t⟹\frac{dx}{2}=dt$

$I=\frac{2}{\sqrt{2}}\int e^t(\sin t+\cos t)dt$

It is in the form of

$\int e^x\left(f\right(x)+f^{{}^{\prime }}(x\left)\right)dx=e^{x}f\left(x\right)+c$

$\therefore I=\sqrt{2}{e}^t\sin t+c$

$=\sqrt{2}{e}^{\frac{x}{2}}\sin \frac{x}{2}+c$ ........$\because t=\frac{x}{2}$

Hence, option 'A' is correct.