Question

Complete the calculation in Example 7 by using the trigonometric identity ${\sin }^2\left(\frac{\theta }{2}\right)=\frac{1}{2}(1-\cos \theta )$to show that ${\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta =4\sqrt{2}$.

Short answer

The integral ${\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta$is equals to$4\sqrt{2}$

Step-by-step solution

Given information

The integral is${\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta$

Calculation

Consider the integral, ${\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta$.

The objective is to solve the integral within the given limits.

By using the trigonometric identity the integral can be written as follows.

Take the integral,

$$\begin{gathered} {\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta ={\int }_0^{2\pi }\sqrt{2{\sin }^2\frac{\theta }{2}}d\theta \left[since{\sin }^2\frac{\theta }{2}=\frac{1}{2}\sqrt{1-\cos \theta }\Rightarrow 2{\sin }^2\frac{\theta }{2}=\sqrt{1-\cos \theta }\right] \\ =\sqrt{2}{\int }_0^{2\pi }\sqrt{{\sin }^2\frac{\theta }{2}}d\theta \\ =\sqrt{2}{\int }_0^{2\pi }\sin \frac{\theta }{2}d\theta \end{gathered}$$

Thus,

$$\begin{gathered} {\int }_0^{2\pi }\sqrt{2{\sin }^2\frac{\theta }{2}}d\theta =\sqrt{2}{\left(-2\cos \frac{\theta }{2}\right)}_0^{2\pi }\left[since{\int }_0^{2\pi }\sin \frac{\theta }{2}d\theta ={\left(-2\cos \frac{\theta }{2}\right)}_0^{2\pi }\right] \\ =\sqrt{2}\left(-2\cos \frac{2\pi }{2}+2\cos \frac{0}{2}\right)\text{By applying the limits} \\ =\sqrt{2}\left(-2\cos \pi +2\cos \frac{0}{2}\right) \\ =\sqrt{2}(-2(-1)+1·2) \end{gathered}$$

On further simplification,

$$\begin{gathered} {\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta =\sqrt{2}(2+2) \\ =\sqrt{2}\times 4 \\ {\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta =4\sqrt{2} \end{gathered}$$

Therefore, the integral ${\int }_0^{2\pi }\sqrt{1-\cos \theta }d\theta$is equals to $4\sqrt{2}$.