Question

Each of the integral expressions that follow represents the area of a region in the plane bounded by a function expressed in polar coordinates. Use the ideas from this section and from Chapter 9 to sketch the regions, and then evaluate each integral

$4{\int }_0^{\frac{\mathrm{\pi }}{4}}CO{S}^{2}2\theta d\theta$

Short answer

The value of the integral is$\frac{\mathrm{\pi }}{2}$square unit

Step-by-step solution

Step 1. Given information

Integral:

$4{\int }_0^{\frac{\mathrm{\pi }}{4}}CO{S}^{2}2\theta d\theta$

 Step 2. Plot the curve 

Since the area of a function is $r=f\left(\theta \right)$is$\frac{1}{2}{\int }_a^b{r}^{2}d\theta$

So by comparing the given integral we get,

$r=2\sqrt{2}\cos 2\theta$

So the graph of this curve is:

Step 3. Evaluate integral

$$\begin{gathered} 4{\int }_0^{\frac{\mathrm{\pi }}{4}}{\cos }^{2}2\theta d\theta \\ =4{\int }_0^{\frac{\mathrm{\pi }}{4}}\frac{1+\cos 4\theta }{2}d\theta \\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}\left(1+\cos 4\theta \right)d\theta \\ =2{\left[\theta +\frac{sin4\theta }{4}\right]}_0^{\frac{\pi }{4}} \\ =2\left[\frac{\mathrm{\pi }}{4}+\frac{\mathrm{sin\pi }}{4}-0\right] \\ =2\left(\frac{\mathrm{\pi }}{4}-0\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$