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Current time:0:00Total duration:5:43

AP.CALC:

FUN‑6 (EU)

, FUN‑6.E (LO)

, FUN‑6.E.1 (EK)

going to do in this video is try to evaluate the definite integral from zero 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 antiderivative here and then evaluate that at PI and then subtract from that it evaluated at 0 so we're probably going to have to use a slightly more sophisticated technique and 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 1 it's a pretty good sign that we should be using integration by parts so let's just remind ourselves about integration by parts so 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 at PI and evaluated 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 u 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 antiderivative so when I take its antiderivative it doesn't get more complicated so 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 it anti-derivative then this expression well 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 so I could write that over here so my f of X I will say is X in which case f prime of X 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 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 the those arbitrary constants are going to get cancelled 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 something 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 it'll make it a little bit simpler sine of X DX 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 it were if I were to say the antiderivative of it we know that the derivative of a sine is negative sine of X and so in fact if we what we want we could bring this negative sign into the integral so we can 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 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 the 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

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