# Can you think of a different way?

## Any way you can calculate this integral ${\int}_{0}^{\frac{\pi}{4}} \left(\sqrt{1 + \cos x}\right) \mathrm{dx}$ without using the trigonometric identity $1 + \cos x = 2 {\cos}^{2} \left(\frac{x}{2}\right)$ ?

Feb 2, 2018

#### Explanation:

$\sqrt{1 + \cos x} = \sqrt{1 + \cos x} \cdot \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}}$

$= \frac{\sqrt{{\sin}^{2} x}}{\sqrt{1 - \cos x}}$

$= \frac{\left\mid \sin x \right\mid}{\sqrt{1 - \cos x}}$

So,

${\int}_{0}^{\frac{\pi}{4}} \sqrt{1 + \cos x} \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} \frac{\sin x}{\sqrt{1 - \cos x}} \mathrm{dx}$

Now use $u = 1 - \cos x$