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# Integration by parts: Example 1

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.