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JH
Numerade Educator

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Problem 40 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \csc^3 x dx $

Answer

$$
\sqrt{3}-\frac{1}{3}-\frac{1}{2} \ln (2 \sqrt{3}-3)
$$

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Video Transcript

this problem is from chapter seven, Section two Problem forty of the book Calculus Early. Transcendental. Lt's a tradition by James Door and we have a definite integral of Kosi can. Cute. So let's start off by doing integration. My parts Let's take you too, because he can So that do you is negative. Cosi again. Time to pay attention, D X and buy a choice if you were left over with TV, which has to be cause he can't you, Ursus is because you can swear, Deeks. So that be is negative attention of books. Okay, so we comply. Are under grace. My parents Were you tansy. So negative. Cozy again. Times call attention here and points are five or six three minus the integral of video notice here were negative on the and also on to you. Those will cancel out. So we just have co attention. Times co tension times co seeking so cozy engine square because he can. So if you think so, observe here. First we see in this integral we have co tension square so we can apply our protagonist identity over here and read on the right to rewrite this as cause he can't squared minus one. So let's go ahead and replace CO. Attention squared with co sequence four minus one. Let's also go to the side and simplify this cause he can cho tension and evaluated at these end points. So let's call this jay So we have J equals negative cause he can't power. Three. Attention Preparatory minus Cho Seon over six. Cho Attention five or six and going in and using our knowledge from the unit circle, we can evaluate each of these to over three one of our route three minus two times three, which we could simplify as two three minus two over three. So that's a J. So that's over here. Circle and green. So picking up where we left off, we have James, which is now two three, minus two or three minus. And now for our new into rule, we have Kosi can square times. Costigan is cozy, can cute, which you would have noticed that's original problem, and then Kosi can X times minus one is negative. Kosi can IX. So since we see the original integral popping up again, let's go ahead and to know this original problem, it's integral by eye so that we see that here. Once again, we have eye again. So's Gordon. Move this eye to the other side. But let's be careful here because we also have a negative Cosi kit and the integral. So we have two I two three minus two over three. And then we have a minus minus. So plus integral Piper six number three Kosi convicts D X. So for the next there, we need to discuss this integral Cosi convicts. So first, let's just fight the previous equation that we had. We have two I two three minus till Rhodri. What now this in a girls in your table's in the section so we can go out and use that fact to evaluate this inaugural. So we have two or three minus two of the three. Then we have minus natural log Absolute value Cho tension of X plus Cassie Innovex evaluated at the end points however six and territory. All right, so this is still too I So then now we're ready tto softer isolates divide both sides by two. So we have We're three minus one over three and then we have minus one half and now we plug in pi over three and five or six. And Tex So plugging in power three first. And then we subtract when we plug in power for six. So here we can go ahead and evaluate. And we have natural log of excuse me, one over. Radical tree, plus two of a radical theory. And again here, um, pushing the minus sign through the apprentices. So that's a plus natural log. And then I have a one happens. Well, radical three plus two. And excuse me. There was, ah matter, mistake here. That should have been a rule three. And we could go on and clean this up a little bit. So here we have Ellen of three over room three bushes, just room three and again using some properties of the logs. You could even simplify this latest expression if you wanted to, but otherwise, this is our final answer.