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$r = \sin^2(\theta/3)$

I can't integrate the following expression:

$$\int \sqrt{\left(\sin^2 \frac x 3\right)^2 + \left(\frac 2 3 \sin \frac x 3 \cos \frac x 3\right)^2}dx$$

I got this in my mid-term and I was wondering what the hell is wrong with the teacher.

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-1 Maybe the problem isn't with the teacher. Who else could be the problem? I mean, teachers do give bad problems sometimes and that is sort of crappy for the students, but for a student to assume they know so much as to know without a doubt that the problem is with the teacher is arrogance. And, if you disrespect your teacher to us, why would we have any incentive to teach you? We know how you treat your teachers. –  Graphth Oct 26 '12 at 16:07
    
He said the integrals would be easy to solve, so we don't spend too much time trying to solve it. –  Gladstone Asder Oct 26 '12 at 16:11
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Well, I figured it out and I didn't think it was all that tough. Perhaps it is a bit harder than what the teacher said, but that's hard for me to know. Maybe if you changed your tone a bit, you might get some help. –  Graphth Oct 26 '12 at 16:17
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1 Answer 1

up vote 4 down vote accepted

The integral is not easy, but doable. I will not worry about constants. Make the change of variable $\phi=\theta/3$. Or not. Bring a $\sin\phi$ outside the square root. (Technically it should be $|\sin\phi|$, but if $\phi$ does not stray beyond $[0,\pi]$ we don't need to worry.) We are integrating $$\sin\phi\sqrt{\sin^2\phi+(4/9)\cos^2\phi}.$$ Inside the square root, replace $\sin^2\phi$ by $1-\cos^2\phi$. So we need to find the integral of $$\sin\phi\sqrt{9-5\cos^2\phi}.$$ Make the substitution $u=\cos\phi$. We end up needing to integrate $$\sqrt{9-5u^2}.$$ Now it is somewhat downhill. Let $u=\dfrac{3}{\sqrt{5}}v$.

With particular limits, things might be easier. For with appropriate limits the definite integral of $\sqrt{9-5u^2}$ will be the perhaps easily found area of a nice part of a circle.

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