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# Integration by parts: ∫x⋅cos(x)dx

## 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. And the question is, well do I assign f of x to x and g prime of x to cosine of x or the other way around? Do I make f of x cosine of x and g prime of x, x? And that 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 where 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, once again, if we take its antiderivative, that sine of x, it's not any more complicated. If we did it the other way around, if we set 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 set g prime of x equaling to x and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning 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-- I'll write it right here-- g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now let's see, given these assumptions, let's see if we can apply this formula. So this has all of this. Let's see, the right-hand side says f of x times g of x. So f of x is 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-- times g of x, times sine of x dx. Now this was a huge simplification. Now I went from trying to solve the antiderivative of x cosine of x to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And of course, we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x sine of x, x times sine of x, minus the antiderivative of this, which is just 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 a 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-- we get our drum roll now-- it's going to be 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. That was pretty interesting.