Trigonometric Integrals Example 3 Part 1 Hi everyone! Welcome back to integralcalc.com. Today we’re going to be doing another trigonometric integral problem. This one is the integral of p/2, this is on the range from 0 to p/2 of x cos of x dx. The first thing that we want to do we’re trying to figure how to take the integral. In this case, this integral is too complicated for me to just take it without altering it in any way. So I start to think, is it partial fractions that use substitution? Is it integration by parts? It’s not a fraction so I rule out partial fractions. So it’s either use substitution or its integration by parts. Use substitution, the derivative of cosine of x this is what I would use for you and the derivative I don’t think would simplify. It’s something that I could cancel so I’m guessing that it’s going to be integration by parts and I have x here as one term then cosine of x is another term. And I like integration by parts because if I assign u to x here then my d is going to be 1 which potentially could simplify things. So I’m going to go ahead and try integration by parts and see if it works. So the integration by parts formula is, remember the integral of udv equals uv minus the integral of vdu. And if you’ve never done integration by parts problem go ahead and checkout the integration by parts section on my website. For this video, I’m going to assume that you have some background knowledge about this so you don’t have to go into too much. Let’s go ahead and apply this formula. What this means is, here’s our original integral and this is what we’re going to be playing into later. Since this is our original integral, we need to assign u and dv to values in our problem. So like I said, I’m going to go ahead and assign u to x, u=x and then I’m going to assign dv to cosine of x. We need to base on u and dv, we need to find du and v. We find du by taking the derivative of u. So the derivative of x is just 1 and then to find v we have to take the integral or the antiderivative of dv. And we’re going to write down another formula here, the integral of cosine of x is actually sine of x. So the integral of this is going to be sine of x. We’ve assigned u and dv into values in our problem and found du and v we can go ahead and plug these four components back into the right hand side of this formula here. So you can see we have u which is x here in our problem and then v which is sine of x and then minus the integral of v which is sine of x and du which is just 1. And then of course we have to add dx to this, dx is just part of our integral notation here. Now let’s go ahead and take away some of these parenthesis and simplify this. We need to make sure that we put the range 0 to p/2 here. We have x sine of x minus the 1 is just going to disappear, it’s redundant. So we just have the integral from 0 to p/2 of sine of x dx. Now let’s go ahead and write down one more formula. The integral of sine of x dx equals negative cosine of x. These formulas here, that’s either something you need to memorize. Something you need to plug in to your calculators that you can reference it on test or write on a formula sheet because you really need to know these if you’re dealing with trigonometric integrals and there’s no really other way better I think than just memorizing, could be integral or sine of x as negative cosine of x so that you are able to take this. So now that we’ve done this we can actually go ahead and take the integral of sine of x. So we’re going to have x sine of x minus, let’s go ahead and just drop because this is the start here of the integral that we’re going to get here. Remember our formula the integral of sine of x is negative cosine of x. So let’s go ahead and write negative cosine of x and then of course we’re suppose to evaluate this entire functions here on the range 0 to p/2 the way that we’re sure of that is we drop big line and we say 0 to p/2. So let’s go ahead and simplify this one more time.