What is the area enclosed by $r=theta^2cos(theta+pi/4)-sin(2theta-pi/12)$ for $theta in [pi/12,pi]$ ?

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Answer 1 · 2017-10-02T15:09:42

The area is about 19.1849 by computational integration.

The area $S$ enclosed by a polar curve $r=f(theta)$ for $theta in [alpha,beta]$ is obtained as follows.

$S=1/2int_alpha^beta{f(theta)}^2d$ $theta$ .
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cited from a Japanese site, 極座標表示の場合の面積 http://www.geocities.jp/phaosmath/int1/polar.htm

Thus, the area in this question is:
$S=1/2int_(pi/12)^(pi){theta^2cos(theta+pi/4)-sin(2theta-pi/12)}^2d$ $theta$ .

This integration is so hard…
According to www.wolframalpha.com $S$ is about 19.1849.
http://www.wolframalpha.com/input/?i=integrate+%7Bx%5E2cos(x%2Bpi%2F4)-sin(2x-pi%2F12)%7D%5E2%2F2+for+x%3D%7Bpi%2F12,pi%7D

What is the area enclosed by r=theta^2cos(theta+pi/4)-sin(2theta-pi/12) r=theta^2cos(theta+pi/4)-sin(2theta-pi/12) for theta in [pi/12,pi]theta in [pi/12,pi]?