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## Integration by parts

Current time:0:00Total duration:7:17

# Integration by parts: ∫𝑒ˣ⋅cos(x)dx

AP.CALC:

FUN‑6 (EU)

, FUN‑6.E (LO)

, FUN‑6.E.1 (EK)

## Video transcript

Let's see if we can use
integration by parts to find the antiderivative of
e to the x cosine of x, dx. And whenever we talk about
integration by parts, we always say, well,
which of these functions-- we're taking a product
of two of these-- which of these functions, either the
x or cosine of x, that if I were to take its derivative,
becomes simpler. And in this case neither
of them become simpler. And neither of them become
dramatically more complicated when I take their
antiderivative. So here, it's kind of a
toss up which one I assign to f of x and which one
I assign to g prime of x. And actually, you can solve
this problem either way. So let's just assign this one. Let's assign f of x
equaling e to the x. And let's assign g prime of
x as equaling cosine of x. So let me write it down. We are saying f of x
is equal to e to the x, or f prime of x is
equal to e to the x. Derivative of e to the
x is just e to the x. And we can say that g-- we're
making the assignment-- g prime of x is equal
to cosine of x. And the antiderivative
of that g of x is also. Or the antiderivative
of cosine of x is just going to be
equal to sine of x. So now let's apply
integration by parts. So this thing is
going to be equal to f of x times g of x,
which is equal to e to the x times sine of x,
minus the antiderivative of f prime of x-- f prime
of x is e to the x. e to the x times g of x, which
is once again, sine of x. Now, it doesn't look like
we've made a lot of progress, now we have an
indefinite integral that involves a sine of x. So let's see if we can
solve that, let's see if we solve this one separately. So let's say if we were trying
to find the antiderivative. The antiderivative of e
to the x, sine of x dx. How could we do that? Well, similarly, we can set f
of x as equal to e to the x. So, and now this is
we're reassigning, although we're happening to make
the exact same reassignment. So we're saying f of x
is equal to e to the x. f prime of x is equal to
just the derivative of that, which is still e to the x. And then we could say
g of x, in this case, is equal to sine of x. We'll put these assignments in
the back of our brain for now. And then-- let me make this
clear-- g prime of x, let me, woops, there you go--
so we have g prime of x is equal to sine
of x, which means that its antiderivative
is negative cosine of x. Derivative of cosine
is negative sine, derivative of negative
cosine is positive sine. So once again, let's apply
integration by parts. So we have f of x times g of x. f of x times g of
x is negative-- is I'll put the negative
out front-- it's negative e to the x times cosine of x,
minus the antiderivative of f prime of xg of x. F prime of x is e to x. And then g of x is
negative cosine of x. So I'll put the cosine
of x right over here, and then the negative,
we can take it out of the integral sign. And so we're
subtracting a negative. That becomes a positive. And of course, we have
our dx right over there. And you might say, Sal, we're
not making any progress. This thing right
over here, we now expressed in terms
of an integral that was our original integral. We've come back full circle. But let's try to do
something interesting. Let's substitute back
this-- all right, let me write it this way. Let's substitute back
this thing up here. Or actually, let me
write it a different way. Let's substitute this for
this in our original equation. And let's see if we got
anything interesting. So what we'll get is
our original integral, on the left hand side here. The indefinite integral
or the antiderivative of e to the x cosine of
x dx is equal to e to the x sine of x, minus
all of this business. So let's just subtract
all of this business. We're subtracting all of this. So if you subtract negative
e to the x cosine of x, it's going to be positive. It's going to be positive
e to the x, cosine of x. And then remember, we're
subtracting all of this. So then we're going to subtract. So then we have minus the
antiderivative of e to the x, cosine of x,dx. Now this is interesting. Just remember all we did is, we
took this part right over here. We said, we used
integration by parts to figure out that it's
the same thing as this. So we substituted this back in. When you subtracted it. When you subtracted
this from this, we got this business
right over here. Now what's interesting
here is we have essentially an equation where we
have our expression, our original expression, twice. We could even assign
this to a variable and essentially solve
for that variable. So why don't we
just add this thing to both sides of the equation? Let me make it clear. Let's just add the integral
of e to the x cosine of x dx to both sides. e to the x, cosine of x, dx. And what do you get? Well, on the left
hand side, you have two times our original integral. e to the x, cosine of x, dx is
equal to all of this business. Is equal to this. I'll copy and paste it. So copy and paste. It's equal to all of that. And then this part, this part
right over here, cancels out. And now we can solve for
our original expression. The antiderivative of e
to the x cosine of x dx. We just have to divide
both sides of this, essentially an equation, by 2. So if you divide the
left hand side by two, you're left with our
original expression. The antiderivative of e
to the x cosine of x dx. And on the right hand side you
have what it must be equal to. e to the x sine of x, plus e
to the x cosine of x over 2. And now we want to be
careful because this is an antiderivative of
our original expression, but it's not the only one. We always have to remember,
even though we've worked hard and we've done-- we've used
integration by parts twice. And we've had to
back substitute in. We have to remember we should
still have a constant here. So if you take the
derivative of this business, regardless of what
the constant is, you will get e to
the x cosine of x. And it's actually a pretty
neat looking expression.