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## Homework Statement

[tex] r = \theta + sin(2\theta)[/tex] for [tex] 0 \le \theta \le \pi [/tex]

a. Find the area bounded by the curve and the x-axis

b. Find the angle [tex]\theta[/tex] that corresponds to the point on the curve with x-coordinate -2

c. For [tex]\frac{\pi}{3} < \theta <\frac{2\pi}{3} , \frac{dr}{d\theta}[/tex] is negative. What does this fact say about r? What does this fact say about the curve?

d. Find the value of [tex]\theta[/tex] in the interval [tex] 0 \le \theta \le \frac{\pi}{2}[/tex] that corresponds to the point on the curve in the first quadrant with the greatest distance from the origin.

## The Attempt at a Solution

a. [tex]1/2\int_{0}^{\pi }(\theta + sin2\theta)^2 d\theta}[/tex]

When I foil I end up with [tex] \theta^2 + 2\theta\sin2\theta + sin^2(2\theta)[/tex]

this is the part where I believe I made the mistake. I know the area is 4.9348 by the use of my calculator...if it is indeed correct...I can easily integrate the 1st and 3rd term...but i think the second term needs to be done by parts if im not mistaken.

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