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
Given:
- Consider the integral {eq}\int\limits_0^\pi {\sqrt {1 - \cos 2x} dx} {/eq} .
Recall the trigonometric identity for {eq}\cos 2x{/eq} as:
{eq}\begin{align*} \cos 2x &= 1 - 2{\sin ^2}x\\ 2{\sin ^2}x &= 1 - \cos 2x \end{align*}{/eq}
Substitute {eq}1 - \cos 2x = 2{\sin ^2}x{/eq} into the integral {eq}\int\limits_0^\pi {\sqrt {1 - \cos 2x} dx} {/eq} and integrate it as follows:
{eq}\begin{align*} \int\limits_0^\pi {\sqrt {1 - \cos 2x} dx} &= \int\limits_0^\pi {\sqrt {2{{\sin }^2}x} dx} \\ &= \int\limits_0^\pi {\sqrt 2 \sin xdx} \\ &= \sqrt 2 \left[ { - \cos x} \right]_0^\pi \\ &= \sqrt 2 \left[ { - \cos \left( \pi \right) - \left( { - \cos \left( 0 \right)} \right)} \right] \end{align*}{/eq}
Further simplify the equation as:
{eq}\begin{align*} \int\limits_0^\pi {\sqrt {1 - \cos 2x} dx} &= \sqrt 2 \left[ { - \left( { - 1} \right) - \left( { - 1} \right)} \right]\\ &= \sqrt 2 \left[ {1 + 1} \right]\\ &= 2\sqrt 2 \end{align*}{/eq}
Thus, the solution of the integral is {eq}2\sqrt 2 {/eq} .
Learn more about this topic:
from
Chapter 13 / Lesson 13Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.