Computing definite integral with trig function $$\int^{2\pi}_0 {\sqrt {1-\cos^{2}x}} \;\mathrm{d}x$$
I am not sure where to even begin... I have tried substituting ${\sin^{2}x}$ under the radical but this does not get me the answer I need.
Any help is appreciated.
 A: Notice, you can also split the limits as follows  $$\int_{0}^{2\pi}\sqrt{1-\cos^2 x}\ dx=\int_{0}^{2\pi}\sqrt{\sin^2 x}\ dx$$
$$=\int_{0}^{2\pi}|\sin x|\ dx$$
$$=\int_{0}^{\pi}|\sin x|\ dx+\int_{\pi}^{2\pi}|\sin x|\ dx$$
$$=\int_{0}^{\pi}\sin x\ dx+\int_{\pi}^{2\pi}(-\sin x)\ dx$$
$$=[-\cos x]_{0}^{\pi}-[-\cos x]_{\pi}^{2\pi}$$
$$=[-\cos \pi+\cos 0]-[-\cos 2\pi+\cos \pi]$$
$$=[1+1]-[-1-1]=2+2=\color{red}{4}$$
A: We have
$$
\int_{0}^{2\pi} \sqrt{1 - \cos^{2}x}dx = \int_{0}^{2\pi}|\sin x| dx = 2\int_{0}^{\pi}\sin x dx.
$$
A: $$\int_{0}^{2\pi}\sqrt{1-\cos^2(x)}\space\text{d}x=$$

Express $\sqrt{1-\cos^2(x)}$ in terms of absolute value:

$$\int_{0}^{2\pi}|\sin(x)|\space\text{d}x=$$

The zeros of $\sin(x)$ in the integration domain are $x=\pi$. Since the sign of $\sin(x)$ is the same between two zeros, integrate over these regions separately and pull out the absolute value:

$$\left|\int_{0}^{\pi}\sin(x)\space\text{d}x\right|+\left|\int_{\pi}^{2\pi}\sin(x)\space\text{d}x\right|=$$
$$\left|\left[-\cos(x)\right]_{0}^{\pi}\right|+\left|\left[-\cos(x)\right]_{\pi}^{2\pi}\right|=$$
$$\left|\left(-\cos(\pi)\right)-\left(-\cos(0)\right)\right|+\left|\left(-\cos(2\pi)\right)-\left(-\cos(\pi)\right)\right|=$$
$$\left|\left(--1\right)-\left(-1\right)\right|+\left|\left(-1\right)-\left(--1\right)\right|=$$
$$\left|1+1\right|+\left|-1-1\right|=$$
$$\left|2\right|+\left|-2\right|=$$
$$2+2=4$$
