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# Integration by parts: ∫x⋅cos(x)dx

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.E (LO)
,
FUN‑6.E.1 (EK)

## Video transcript

in the last video I claimed that this formula would come handy for solving or for figuring out the antiderivative of a class of functions let's see if that really is the case so let's say I want to take the antiderivative of x times cosine of X DX now if you look at this formula right over here you want to assign part of this to f of X and some part of it to G prime of X so the question is well do I sign f of X 2x and G prime of X 2 cosine of X or the other way around do I make f of X cosine of X and G prime of X X and the thing to realize is to look at the other part of the formula and realize that you're essentially going to have to solve this right over here and here we have the derivative of f of X times G of X so what you want to do is assign f of X so that the derivative of f of X is actually simpler than f of X and assign G prime of X that if you were to take its antiderivative it doesn't really become any more complicated so in this case if we assign f of X to be equal to X F prime of X is definitely simpler F prime of X is equal to 1 if we assign G prime of X to be cosine of X to be cosine of X once again if we take its antiderivative that's sine of X it's not any more complicated if we did it the other way around if we said f of X to be cosine of X then we're taking its derivative here that's not that much more complicated but if we said G prime of X equaling 2x and then we had to take its antiderivative you get x squared over 2 that is more complicated so let me make it clear over here we are assigning f of X f of X to be equal to X and that means that the derivative of F is going to be equal to 1 we are assigning we are assigning alright right here G prime of X to be equal to cosine of X which means G of X is equal to sine is equal to sine of X the antiderivative of cosine of X now let's see given the this these assumptions let's see if we can apply this formula so this has all of this let's see let's the right-hand side says f of X times G of X so f of X is X f of X is X G of X is sine of X G of X is sine of X and then from that we are going to subtract the antiderivative of f prime of X well that's just 1 that's just 1 times G of x times sine of X x times sine of X DX now this was a huge simplification now I just have to I went from trying to solve the antiderivative of X cosine of X 2 now I just have to find the antiderivative of sine of X and we know the antiderivative of sine of X we know the antiderivative of sine of X DX is just equal to negative cosine of X and so and of course we can throw the plus C and now now that we're pretty done with taking all of our anti derivatives so this is all of this is going to be equal to X sine of X x times sine of X sine of X minus minus the antiderivative of this which is just negative cosine of X minus negative cosine of X and then we could throw in a plus C right at the end of it and doesn't matter if we subtract 2 C or add the C we're saying this is some arbitrary constant which could even be negative and so this is all going to be equal to just do we get our drumroll now it's going to be x times sine of X x times sine of X subtract a negative that becomes a positive plus cosine of X plus C and we are done we were able to take the antiderivative of something that we didn't know how to take the antiderivative of before and that was that was pretty interesting
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