# How do you Integrate ?

## ${\int}_{0}^{2 x} \sqrt{1 + \sin \left(\frac{x}{2}\right)} \mathrm{dx}$

May 4, 2018

$\sin \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2}\right) - 1$

#### Explanation:

${\int}_{0}^{2 x} \sqrt{1 + \sin \left(\frac{x}{2}\right)} \cdot \mathrm{dx}$

color(green)(sin(x/2)=2sin(x/4)cos(x/4)

color(green)(1=sin^2(x/4)+cos^2(x/4)

Substitute

${\int}_{0}^{2 x} \sqrt{1 + \sin \left(\frac{x}{2}\right)} \cdot \mathrm{dx}$

$= {\int}_{0}^{2 x} \sqrt{{\sin}^{2} \left(\frac{x}{4}\right) + 2 \sin \left(\frac{x}{4}\right) \cos \left(\frac{x}{4}\right) + {\cos}^{2} \left(\frac{x}{4}\right)} \cdot \mathrm{dx}$

By Completing The Square

$= {\int}_{0}^{2 x} \sqrt{{\left(\sin \left(\frac{x}{4}\right) + \cos \left(\frac{x}{4}\right)\right)}^{2}} \cdot \mathrm{dx}$

$= {\int}_{0}^{2 x} \left(\sin \left(\frac{x}{4}\right) + \cos \left(\frac{x}{4}\right)\right) \cdot \mathrm{dx}$

$= {\left[\sin \left(\frac{x}{4}\right) + \cos \left(\frac{x}{4}\right)\right]}_{0}^{2 x}$

$= \sin \left(\frac{x}{2}\right) + \cos \left(\frac{x}{2}\right) - 1$

color(blue)("if this question was like this " sqrt(1+cos(x/2)) " it can be simplified to look like this"

color(blue)(sqrt2*cos(x/4) "using the Double angles Formulae"