Question

Evaluate

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

Answer 1

We have,

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

Using Identity

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

Now,

$I={\int }_0^{2\pi }{e}^x(\cos x\cos \frac{\pi }{4}-\sin x\sin \frac{\pi }{4})dx$

$={\int }_0^{2\pi }{e}^x(\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x)dx$

Now, let $f\left(x\right)=\frac{\cos x}{\sqrt{2}}$ then,$f^{\prime }\left(x\right)=-\frac{\sin x}{\sqrt{2}}$

Applying property,

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

Now, we get,

$I={{\left[e^x\frac{\cos x}{\sqrt{2}}\right]}_0}^{2\pi }$

$I=[e^{2\pi }\frac{\cos 2\pi }{\sqrt{2}}-e^0\frac{\cos 0}{\sqrt{2}}]$

$I=[\frac{e^{2\pi }\times 1}{\sqrt{2}}-\frac{1\times 1}{\sqrt{2}}]$

$I=\frac{e^{2\pi }-1}{\sqrt{2}}$

Hence, this is the answer.