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Evaluate the integral.

$ \displaystyle \int_0^{2\pi} \sin^2 \left(\frac{1}{3} \theta \right) d \theta $

$\pi+\frac{3 \sqrt{3}}{8}$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Integration Techniques

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this problem is from Chapter seven, Section two. Problem number eight in the book Calculus Early. Transcendental lt's a Definition by James Store. Here we have a definite overall from zero to pi. Sign square of data over three with respect that data. So for this problem, since we haven't even power on the sign and there's no co signs present, it's going to be best to use this trick in a metric identity, which says Sign square of X is one minus coastline of two x divided by two. So whatever the input is on, the sign over here in co sign multiply the same thing by two. So let's use this time to rewrite this. Integral as the integral from Sierra two pyre, they have a one half minus co sign of two times one third data. So we have ah to that over three. Also, divide this turn by, too, because we're dividing by two over here and now we can evaluate you should these integral separately. So the interval of one half with respect the data simply they divided by two. And for this integral over here we can apply you, so let's take you to be to date over three. So using this use of been evaluating this underworld the second integral we'LL tend sign of to date over three times three over too. And we're evaluating thes this expression at zero in super. So let's plug into my first for data thirteen two pi divided by two minus sign of four pi over three. And then here Let's simplify this fractions. We have three over two and then we have another two on the bottom. So let's just multiply this by three or four minus. Now we plug in Seo for data. So having zero over too, minus sign of zero over two times three over. Horrible. Did you write that? So here we have zero and from trigonometry with no sign of zero zero. So here the twos, cancel. We just have a pyre. The level minus and in sign of four pi over three is negative. Rule three and divide that by two. And then we still have three over four. So here we could just multiply these denominators together to get eight on the bottom and cancel out this double negative tub tin PIDE plus three, Route three divided by. And that's your final answer

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