Integration by parts: definite integrals | AP Calculus BC | Khan Academy

Going to do in this video is try to evaluate the definite integral from 0 to pi of x cosine of x dx. Like always, pause this video and see if you can evaluate it yourself.

Well, when you immediately look at this, it's not obvious how you just straight up take the anti-derivative here and then evaluate that at pi and then subtract from that it evaluated at zero. So we're probably going to have to use a slightly more sophisticated technique.

In general, if you see a product of functions right over here and if one of these functions is fairly straightforward to take the antiderivative of without making it more complicated like cosine of x, and another of the functions like x, if you were to take its derivative, it gets simpler—in this case, it would just become one. It's a pretty good sign that we should be using integration by parts.

So let's just remind ourselves about integration by parts. Integration by parts, I'll do it right over here. If I have the integral, and I'll just write this as an indefinite integral, but here we want to take the indefinite integral and then evaluate it at pi and evaluate it at 0.

So if I have f of x times g prime of x dx, this is going to be equal to—and in other videos we proved this; it really just comes straight out of the product rule that you learned in differential calculus—this is going to be equal to f of x times g of x minus—you then swap these around—minus f prime of x g of x dx.

And just to reiterate what I said before, you want to find an f of x that when I take its derivative, it simplifies it. So simplify, and you want to find a g prime of x that when I take its anti-derivative, it doesn't get more complicated. Not more complicated. Because if the f of x gets simplified when I take its derivative and the g prime of x does not get more complicated when I take its antiderivative, then this expression will maybe be easier to find the antiderivative of.

So let's do that over here. Between x and cosine of x, which one gets more simple when I take its derivative? Well, the derivative of x is just 1. So I'm going to make that my f of x. I could write that over here, so my f of x I will say is x, in which case f prime of x is going to be equal to 1.

And then what would my g prime of x be? Well, my g prime of x is cosine of x. If I take its antiderivative, it doesn't get more complicated. The antiderivative of cosine of x is sine of x. So let me make that my g prime of x. So g prime of x is equal to cosine of x, in which case g of x—the antiderivative of cosine of x—well, it's just sine of x, or another way to think about it: the derivative of sine of x is cosine of x.

Now, you could think about plus c's and all of that, but remember this is going to be a definite integral, so all those arbitrary constants are going to get canceled out.

So now, let's think through this. Let's just apply the integration by parts here. In this particular case, all of this is going to be equal to—so I'm saying that is equal to this. I'm going to skip down here. It's going to be equal to f of x times g of x. So that is f of x is x, g of x is sine of x: f of x times g of x minus the integral of f prime of x.

f prime of x is just 1. We could write like that: 1 times g of x. g of x is sine of x. So we could write it like this, but 1 times sine of x—well, we could just rewrite that as sine of x dx.

And then remember this is a definite integral, so we are going to want to evaluate this whole thing at pi and at 0 and then take the difference between the two. But what is the indefinite integral of sine of x dx?

Well, if I were to say the antiderivative of it, we know that the derivative of cosine is negative sine of x. So, in fact, if we what we want, we could bring this negative sign into the integral. So we could say plus the integral of negative sine of x.

Now, this clearly—the antiderivative here is cosine of x. So this thing is going to be cosine of x, and now we just have to evaluate it at the endpoints.

So let's first evaluate this whole thing at pi. So this is going to be equal to pi sine of pi plus cosine of pi, and then from that, I'm going to subtract this whole thing evaluated at zero. So let me do zero in a different color. At zero, so it's going to be zero times sine of zero plus cosine of zero.

So let's see, sine of pi is just zero, so this is just going to cancel out. Cosine of pi, that is negative one, and then this is zero, and then cosine of zero, that is one. So you have negative one minus one, so this all gets us to negative two, and we are done.

Using integration by parts, we were able to evaluate this definite integral.