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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 .

Answer and Explanation: 1

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Given:


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


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

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Integration Problems in Calculus: Solutions & Examples

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Chapter 13 / Lesson 13
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Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.


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