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Problem 4 Medium Difficulty

Use the indicated entry in the Table of Integrals on the Reference Pages to evaluate the integral.

$ \displaystyle \int_0^1 \tan^3 \left (\frac{\pi x}{6} \right)\ dx $ ; entry 69


$=\frac{1}{\pi}+\frac{6}{\pi} \ln \left(\frac{\sqrt{3}}{2}\right)$


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Video Transcript

Okay. This question wants us to evaluate this integral using a formula from the tape. So if we look at our integral table in the back of the book, it says tow. Look at number 69 which says, but the integral of tangent. Cute. Do you is equal to 1/2 tangent squared of you Plus Ellen. Of course I knew plus c, of course. So since we know this formula, what we want to do is turn original integral into that form. So let's make a U substitution let you equal. Hi, ex over six. So d'you is just pi over six d x or D X equals six over pi, do you? And now we're almost done. But since we have a definite integral, we have to transform our limits. So you of one equals pi times one divided by six So high over six and you have zero is pi times zero over six so still zero Okay, so now we're ready to transform arena girl. So now we have this new expression are integral now goes from zero to pi over six of tangent cubed you do you with our general factor off six over pie in front. So now we just evaluate this using our formula from the integral table. So we get 1/2 tangents squared of you Plus Ln of co sign you and we're evaluating this anti derivative from zero to pi over six. So if we plug those in, we get six over pi times 1/2 tangent squared pi over six plus Ellen of co sign of pirates six minus six over pi times 1/2 tangents squared zero plus Ellen of co sign of zero and tangent of 00 co Sign of zeros one and Ellen of 10 So this term vanishes. So then we're just left with six over pi times 1/2 tangent squared of pirate or six plus Ellen of co sign of pi over six and evaluating these we get six over pile front times. Well, tangents squared of pi over six is group three over three quantity squared plus Ln of route three over two. And what is this? Simplify too. Well, after multiplying through by the six over pie, we get one over pi plus six over pi Ln of route three over to as our final answer

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