💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!



Numerade Educator



Problem 36 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cot^3 x dx $


$\frac{1}{2}+\ln (\sqrt{2} / 2)$


You must be signed in to discuss.

Video Transcript

This problem is from Chapter seven section to problem number thirty six in the book Calculus Early Transcendental Sze eighth edition by James Store Here we have a definite integral from power for so pi over to of course, engine cute. So using the definition of co tension is co sign over sign We can rewrite this integral soco tension cute becomes cosign cubed over Thank you Further weaken Rewrite Cho Sang Cube So let's go to the side and simplify this expression So we're dealing with the numerator right now. So costa and cubed we can rewrite it is coastline square times goes on and using the protectorate identity we can rewrite co signed square is one minus sine squared So if we apply this identity to the numerator we can re read on a rule So in the numerator we have one minus science where times call center books In our denominator, we could just leave It is thank you because it looks like at this point, we could basically using u substitution. So there's no no need toe rewrite the denominator. So at this point, it looks like we can do a u substitution where we can take you Just be Santa books If we're going to do this well first, let's note that to us Cose, Innovex, DX. But because we're doing a definite and rule, we should also be prepared to change these limits of integration. So the new bounds of integration We have the lower end points that a popover four will be signed. A pirate for let's from the unit circle. He's wearing two over too. In the upper end point instead of Piper too will become sign a pirate too, which is one So those are new limits of integration for the new integral. So let's come over here and toe continue where we left off in the integration after our use of there are new limits of integration. Groom too over two to one and then we have Ah one minus you square overyou cubed Deal. Let's split up this fraction. So we have ah one over you. Cube, which weaken right is you to the minus three minus. You squared over you cute. So one over you, which we can raise you to the minus one so we can use the power rule to evaluate each of these intervals. So for the first interval that becomes you to the minus two over Negative too. And for the second, we have minus Ellen absolute value of you and our end points are still group two over till and one. So let's plug these in for for you. So when I plug in one first for you, we just have a one over negative too. Minus Ellen of one. Then when we plug in, grew to over too. We have so first we can rewrite you to the minus two is one over you square. So, doing that, we have one over negative two times you squared which will be rude too. Over to squared, minus Elland growing to over two. So here we have naturalized of one, which is zero. So that goes away. We have a negative one half the's minuses will cancel out that becomes a plus one over two. And if we square this parentheses over here, we get two over four. So on the numerator and in the denominator, we have ah two times two over four. So that becomes a one over one, which is one and then we also have another minus and a minus off another plus time's a natural log and really here we can. We can drop the absolute value because route to over two is positive. So we have positive one and then a minus one half. So that's a positive one half. And then we have this Ellen's room left over at the end, and there's our answer.