Question

Can someone please help me! I am trying to integrate r=1+cos^2(2theta) to find the polar area but I just don't know how to start!

Answer 1

@Directrix

Answer 2

If you want the entire bounded region's area:
\[\begin{align*}A&=\iint_R dA\\\\
&=\int_0^{2\pi}\int_0^{1+\cos^22\theta}r~dr~d\theta
\end{align*}\]
or simply
\[A=\frac{1}{2}\int_0^{2\pi}\left(1+\cos^22\theta\right)^2~d\theta\]

Answer 3

For the integral, I suggest expanding and making use of this identity:
\[\cos^2x=\frac{1+\cos2x}{2}\]

Answer 4

how do you integrate this tho? that is where I am having trouble \[1/2\int\limits_{0}^{2\pi} (1+\cos^2(2\theta))^2 d(\theta)\]

Answer 5

First expand:
\[(1+\cos^22\theta)^2=1+2\cos^22\theta+\cos^42\theta\]
From the identity you have
\[\begin{align*}(1+\cos^22\theta)^2&=1+2\frac{1+\cos4\theta}{2}+\left(\cos^22\theta\right)^2\\\\
&=2+\cos4\theta+\left(\cos^22\theta\right)^2\\\\
&=2+\cos4\theta+\left(\frac{1+\cos4\theta}{2}\right)^2\\\\
&=2+\cos4\theta+\frac{1}{4}\left(1+2\cos4\theta+\cos^24\theta\right)\\\\
&=2+\cos4\theta+\left(\frac{1}{4}+\frac{1}{2}\cos4\theta+\frac{1}{4}\left(\frac{1+\cos8\theta}{2}\right)\right)\\\\
&=2+\cos4\theta+\left(\frac{1}{4}+\frac{1}{2}\cos4\theta+\frac{1}{8}+\frac{1}{8}\cos8\theta\right)\\\\
&=\frac{19}{8}+\frac{3}{2}\cos4\theta+\frac{1}{8}\cos8\theta
\end{align*}\]