- [Voiceover] Let's say that we want to evaluate
this indefinite integral right over here.
And you immediately say hey, you've got the square root
of four mins X squared in the denominator,
you could try to use substitution,
but it really doesn't simplify this in any
reasonable way.
How do you tackle this?
And an insight that you might have is well
this thing, this square root of four minus X squared,
this looks like something that I might get
if I'm dealing with the Pythagorean theorem,
if I'm solving for non-hypotenuse sides
of the Pythagorean theorem,
especially if the length of the hypotenuse
is two, this would be two squared
and then the other side is X,
then two squared minus X squared,
this would be the length of the other side.
Well let's just run with that then.
So, if we have,
if we have a,
and if any of you didn't feel that inside immediately,
it's okay.
Because to be honest, I actually did not feel that
inside probably the first time I saw things like this.
So let's say you have a right triangle.
So let's just visualize that insight that we talked about.
So if the hypotenuse of the right triangle is of length two,
the hypotenuse is of length two,
hypotenuse is of length two
and let's say this side, right over here.
Actually, let me do this side.
Let's say this side...
Actually, I'm gonna go with this side.
This side right over here is of length X.
It's a little counter intuitive, because normally
we associate this side over here with Y,
but let's just go with it.
This side right over here is of length X.
Then what would this side right over here be?
What would this other non-hypotenuse length be?
Well, just if you solve for it using the Pythagorean Theorem
you get that it is going to be
the square root of the hypotenuse squared,
two squared, which is just four,
minus the other side squared.
so minus X squared.
Well that's interesting, that is this expression
that insight, that intuition that we
may have had when we saw this right over here,
but still, how does that help us?
Well this is where the trigonometry comes in.
Because if we define,
if this angle right over here,
we say this is theta,
then what is sine and cosine of theta going to be
in terms of these sides?
Well let's see,
the sine of theta,
sine of theta is equal to the opposite
over the hypotenuse.
Is equal to X over two.
Or if you want to solve for X,
we get X is equal to two sine theta.
Well that's interesting.
What about the cosine of theta?
Cosine of theta is equal to the adjacent side,
square root of four minus X squared
over the hypotenuse.
Or if you want to solve for this side,
it's going to be,
we could say that the square root of four
minus X squared is going to be equal to
the hypotenuse times the cosine of theta.
So that's interesting.
If X is equal to two sine theta,
then this other side, then this entire expression
simplifies to two cosine theta.
And that seems pretty interesting now.
So let's make the substitution.
Let's say that X is equal to two sine theta.
And if X is equal to two sine theta,
then DX is going to be equal to
two cosine theta D theta.
And then if we have X as equal to two sine theta,
then what is this thing right over here?
Well we just figured out,
this thing is two cosine theta.
This thing is equal to two,
actually, let me do it in that orange color.
Two cosine theta.
This is equal to two cosine theta.
And we were able to do this,
drawing this right triangle and using
the sohcahtoa definition of these trig functions,
and obviously we could use units circle,
that's kind of an extension of these.
But you could also do it if you say
hey, look you know, if this is two sine theta
you could use the Pythagorean identity,
our trigonometric identities and you would
see that this entire expression,
if you did it that way, would simplify to two cosine theta.
But now let's just run forward
and let's see if we can evaluate it
using this substitution.
So this is going to be the indefinite integral,
so DX, so one times DX, that's going to be
DX is two cosine theta D theta.
So let me right that, that is
two cosine theta
D, I'll right the D theta out here,
and then what's the denominator?
Square root of four minus X squared?
Well that's two cosine theta again.
So that is two
cosine theta.
Well this seems to work out quite nicely.
If you two cosine theta over two cosine theta,
that's just going to be one.
This simplifies.
This simplifies to
D theta, which is, if you just evaluate this,
this is just going to be equal to
theta plus C.
Well, this is kind of nice, but we're still not done.
We want our indefinite integral
in terms of X.
So now let's just solve for X here.
So if X is two cosine theta,
sorry, if X is two sine theta,
so X is equal to two sine,
X is equal to two sine theta,
then let's see, divide both sides by two.
X over two is equal to sine theta.
And then if you want to solve for theta,
theta's the angle that if you take the sine of it
you get X over two.
So we could say,
let's give ourselves a little bit more real estate,
that theta
is equal to the inverse sine,
the inverse sine of this thing,
X over two.
We could write it that way,
or we could write that theta
is equal to the arcsine.
The arcsine of,
the arcsine of X over two.
So this is going to be equal to the,
theta is arcsine of X over two,
so let's just do that.
Arcsine of X over two
plus C.
And we're done.
We've just evaluated that indefinite integral.
Now some of you might have noticed something.
I kind of brushed past it really fast
just to give you the big,
to see the forest for the trees,
but there's some interesting details
and it's, I think, worth digging
into a little bit over here.
So the first one is, you might see that look
the restriction on X here is,
or the domain here on this expression
is restricted.
So let's just keep track of that
to make sure that we didn't do anything strange
when we did that substitution.
So the domain here, X has to be
X has to be greater than negative two
and less than two.
If the absolute value of X were equal to two,
then you would have a zero in the denominator.
If the absolute value of X is greater than two,
then you're gonna have a negative in the denominator
and that's not defined.
So this right over here is the domain.
So let's make sure that our substitution
didn't do anything weird with that.
So if X has to be between negative two and two,
and we're saying X is two sine theta,
that means two sine theta would have to be
between negative two and two.
So negative two would have to be less than
two sine theta.
Two sine theta.
Which would have to be less than two.
We could divide all the different parts of this
compound inequality by two
and you're gonna get negative one is less than sine theta.
Is less than sine theta,
which is less than one.
And the way we can do that is if theta
is less than pi over two.
At pi over two, sine of theta would be equal to one.
And if theta is greater than negative pi over two.
So if we restrict it in this way,
if we say theta is going to be
in this range right over here,
then we are restricting our domain in a reasonable way.
And this works out well because this typically is
the range for the arcsine function.
So we could feel good about that.
Now another question that you might have
is okay well look, you know we divided
by cosine of theta here, but that's okay
as long as cosine of theta does not equal zero,
'cause you don't want a zero in the denominator.
And the good thing about this restriction on theta
is as long as theta is greater than
negative pi over two and less than pi over two,
cosine of theta is going to be,
it's not going to be zero, and actually
it's going to be positive.
If negative pi over two or pi over two were allowed,
then you would get a zero down here
and we would have to think about
restricting things in some other ways.
So it looks like everything is cool.
We dug a little deeper.
We said, okay we haven't done anything
strange to the domain or to,
unrestricted it some strange way,
and so we can feel,
we can feel good about,
we can feel good about this,
about this,
this answer that we've gotten.