We now have a definite integral going
from x equals 0 to x equals square root of pi,
of x sine of x squared dx.
So the first thing that you might want to ask yourself
is whether u substitution is appropriate here.
And if it is, what would you set u to be equal to?
I'll let you think about that for a second.
Well, we see here this expression, x squared.
And if we were to just take the derivative of x squared,
it would be 2x.
We have an x here, so it's almost a 2x.
To turn this x into 2x we could multiply it by 2.
But we can't just only multiply it by 2,
because that would change the value
of this entire expression.
We'd also have to multiply by 1/2.
So all I've done so far is I multiplied by 2 and by 1/2,
which is the same thing as multiplying by 1.
I have not changed the value of this entire definite integral.
Now, let's do our u substitution.
If we set u as equal to x squared,
we now know that du is going to be equal to 2xdx.
And so, what can we see here?
Well, we have our u there, that's how we defined it.
And our du is going to be 2x times dx.
So this entire inside, or the thing
that we're taking the integral of, the definite integral of,
is going to be-- we can rewrite it as sine of x squared.
But x squared is just u, so sine of u.
And instead of 2xdx, we know that 2xdx is now du.
So the thing that we're taking the integral of in terms of u
could be expressed in this way.
And so we have our integral.
But these boundaries of integration
were written in terms of x.
We could have written this as x is equal to 0,
to x is equal to square root of pi.
Now we want to write them in terms of u.
And let me not forget that I had this 1/2 sitting out here.
So how do we write it in terms of u?
So when x is equal to 0, what is u going to be equal to?
This boundary is when x equals 0.
Well, when x is equal to 0, u is also equal to 0.
So this is from u equals zero, this is the same boundary.
When x is equal to 0, u is equal to 0,
based on how we defined u.
Now what about here?
What is u equal to when x is equal to the square root of pi?
Well, u is equal to the square root of pi
squared, let me write this down.
u is going to be equal to the square root of pi squared, when
x is equal to the square root of pi, which is just equal to pi.
So this is going to be from u equals 0, to u equals pi.
I didn't have to explicitly write this u equals here,
but that makes it a little bit clearer.
Our boundaries are now written in terms of u.
And so we are now ready to evaluate
this indefinite integral.
So this whole thing is going to be equal to 1/2 times
the expression.
Well, what's the antiderivative of sine of u?
Well, the derivative of cosine of u,
let me write it over here.
ddx-- or maybe I should write in terms of u--
d/du cosine u is equal to negative sine, negative sine
of u.
So the derivative of negative cosine of u
is equal to sine of u.
So the antiderivative of sine of u is negative cosine of u.
And we're going to evaluate that from, at u equals pi.
And from that we're going to subtract it, evaluate it at u
equals 0.
u equals 0.
So this is going to be equal to 1/2 times--
well, what happens when we evaluate it at u equals pi?
Negative cosine of pi.
Cosine of pi, if we just-- and I always
have to redraw my unit circles here-- so cosine of pi.
We're at this point on the unit circle.
Cosine is the x-coordinate.
Cosine of pi is-- let me just write it down--
it's going to be negative cosine of pi.
And then from that, we are going to subtract
this evaluated when u equals 0.
So minus negative cosine of 0.
Not cosine of theta.
That's what it's going to be equal to.
And now we're ready to evaluate it.
Cosine of pi is -1.
But then we have this negative out front,
so this whole thing will evaluate to 1.
Cosine of 0.
Cosine of 0 is 1.
And then we're subtracting a -1, which
is the exact same thing as adding a 1.
So this whole thing also becomes a 1.
And so in here we have 1 plus 1 times 1/2.
So this whole thing simplifies to-- ready for our drum
roll-- 1.
And you see, when we did u substitution
with a definite integral, as long
as we changed our boundaries in terms of u,
we didn't have to go and then undo the substitution.
We didn't even have to go and convert back into x.
We were able to evaluate the definite integral
in terms of u.