Q. 16. - Calculus

Question

complete Example 4 by evaluating the integral

${\int }_0^{2\pi }\sqrt{2+2\cos \theta }d\theta$

Step-by-Step Solution

Answer

The required answer is $8$

Step 1: Given information

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

Step 2: The objective is to find the value of the integral.

The integral ${\int }_0^{2\pi }\sqrt{2+2\cos \theta }d\theta .$

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

Then,

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

Step 3: Find the value of integral

In this case, the supplied function has a negative value in the interval $\pi$ to $2\pi$

Calculate the integral in the range of $0$ to $\pi$ and multiply it by $2$

$$\begin{gathered} {\int }_0^{2\pi }\sqrt{2+2\cos \theta }d\theta =2\left[2{{\left(2\sin \frac{\theta }{2}\right)}_0}^{\pi }\right] \\ =8(\sin \frac{\pi }{2}-0) \\ {\int }_0^{2\pi }\sqrt{2+2\cos \theta }d\theta =8 \end{gathered}$$

Hence, the value of the integral is $8$