- [Instructor] We're gonna 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 anti-derivative here and then evaluate that
at pi and then subtract from that
and evaluate it at zero, 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 anti-derivative 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 one.
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 wanna take
the indefinite integral and then evaluate it at pi
and evaluate it at zero, 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 prove this, it really just comes straight
out of the product rule that you learned
in differential calculus, this is gonna be equal
to f of x times g of x minus, you then swap these around,
minus f prime of x, g of x, dx.
And just to reiterate what I said before,
you wanna find an f of x that when I take its derivative,
it simplifies it, so simplify,
and you wanna find a g prime of x that when I take
its anti-derivative, so when I take its anti-derivative,
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 its anti-derivative,
then this expression will maybe be easier to find
the anti-derivative 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 one, so I'm gonna 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 one,
and then what would my g prime of x be?
Well my g prime of x, cosine of x if I take
its anti-derivative, it doesn't get more complicated.
The anti-derivative 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 anti-derivative 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 gonna be a definite integral
so all of those arbitrary constants are going to get
canceled 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, so I'm saying that is equal to this,
I'm gonna 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 one,
we could write it like that, one times g of x,
g of x is sine of x, so we could write it like this,
but one 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, and then remember, this is a definite
integral, so we are going to want to evaluate
this whole thing at pi and at zero, and then take
the difference between the two.
But what is the indefinite integral of sine of x dx?
Well, or to say the anti-derivative of it,
we know that the derivative of cosine is negative sine
of x, and so in fact what we want, we could bring
this negative sine into the integral, so we could say
plus the integral of negative sine of x,
now this clearly the anti-derivative 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 end points.
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 let's see,
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.