Question

$\int e^{x/2}\sin (\frac{x}{2}+\frac{\pi }{4})dx$ is equal to---

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

Answer 1

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

$\begin{array}{c}\int e^{\frac{x}{2}}.\sin \left(\frac{x}{2}+\frac{\pi }{4}\right)dx\\ let\left(\frac{x}{2}+\frac{\pi }{4}\right)=t\\ \frac{dx}{2}=dt\\ dx=2dt\\ \Rightarrow 2\int e^{t-\frac{\pi }{4}}\sin t\\ \Rightarrow \frac{2}{e^{\frac{\pi }{4}}}\int e^t\sin t=\sin t.\int e^{t}dt-\int \left[\frac{d\sin t}{dt}\int e^{t}dt\right]dt\\ \Rightarrow \frac{2}{e^{\frac{\pi }{4}}}\int e^t\sin t=e^t\sin t+\int \cos t.e^{t}dt\\ \Rightarrow \frac{2}{e^{\frac{\pi }{4}}}\int e^t\sin t=e^t\sin t+\cos t\int e^{t}dt-\int \frac{d\cos t}{dt}\int e^{t}dt]dt\\ \Rightarrow \frac{2}{e^{\frac{\pi }{4}}}\int e^t\sin t=e^t\sin t+\cos t.e^t-\int \sin t.e^{t}dt\\ \Rightarrow \int e^t\sin t\left[\frac{2}{e^{\frac{\pi }{4}}}+1\right]=e^t\left(\sin t+\cos t\right)+c\\ \Rightarrow \int e^t\sin t=\frac{\sqrt{2}{e}^t\sin \left(t+\frac{\pi }{4}\right)\times e^{\frac{\pi }{4}}}{\left(2+e^{\frac{\pi }{4}}\right)}+c\\ =\frac{\sqrt{2}{e}^{\left(\frac{x}{2}=\frac{\pi }{2}\right)}{\sin }^{\left(\frac{x}{2}+\frac{\pi }{2}\right)}}{2+e^{\frac{\pi }{4}}}.\end{array}$