# Evaluate {eq}\displaystyle \int_0^{\pi} \sqrt { 1 - \cos 2x} \ dx {/eq}

## Question:

Evaluate

{eq}\displaystyle \int_0^{\pi} \sqrt { 1 - \cos 2x} \ dx {/eq}

## Integration:

To understand the process of integration, let us consider a simple example that is {eq}\int {{y^{n - 1}}dy} {/eq} .

Now integrating the integral {eq}\int {{y^{n - 1}}dy} {/eq} gives {eq}\frac{{{y^{n - 1 + 1}}}}{{n - 1 + 1}} = \frac{{{y^n}}}{n} + C{/eq} , where {eq}C{/eq} is the arbitrary constant of the indefinite integration and the power of the function is increased .

Become a Study.com member to unlock this answer!

Given:

• Consider the integral {eq}\int\limits_0^\pi {\sqrt {1 - \cos 2x} dx} {/eq} .

Recall the trigonometric identity for {eq}\cos 2x{/eq}...