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Current time:0:00Total duration:5:08

Let's see if we can evaluate
this indefinite integral. And the clue that
trig substitution might be appropriate is
what we see right over here in the denominator
under the radical. In general, if you see
something of the form a squared minus x squared, it
tends to be a pretty good idea, not always, but it's a good clue
that it might be a good idea to make the substitution x
is equal to a sine theta. Because if you do
that, then this will become a squared
minus a squared sine theta. And if you factor
out the a squared, you can start leveraging-- this
is a squared-- you can leverage one of the most
basic trig identities that this right over here
is cosine squared theta and maybe simplify
the expression. Now, you're probably
saying, well, this 8 minus 2x squared,
it's not as obvious that it's a squared
minus x squared. But we could simplify this. Or I guess we can
write it in a way that it starts to
have this pattern. You can rewrite 8 minus 2--
let me write it right under it. You could write 8
minus 2x squared as, if we factor out a 2, as
2 times 4 minus x squared. And now this very clearly
has a pattern a squared minus x squared. You could write this as 2 times
2 squared minus x squared. So in this case, a
would be equal to 2. So let's make that substitution. Let's make the
substitution that x is going to be equal
to 2 sine theta and dx is going to be equal to
2 cosine theta d theta. So what's this part under
the expression going to be? Well, we already started
simplifying it right over here. It's going to become 2 times
2 squared minus x squared. x squared is 2 sine
theta, so x squared is going to be 2 squared
sine theta squared. And now we can factor
out the 2 squared. So this is going to
be 2 times 2 squared times 1 minus sine
squared theta. 2 times 2 squared, well
that's just going to be 8, times cosine squared theta. That's what we have
under the radical. So let's do that. Let's rewrite this
thing up here. So we're going to
have-- and I'll take the pi outside
of the integration. So we're going to
have pi times dx. dx is 2 cosine theta d theta. So it's going to be--
let me make it clear. So dx-- I want to
do that in blue. dx right over there is
2 cosine theta d theta. So let me write
that, 2 cosine theta. And I'll write the
d theta out here. I could have written
it in the numerator. And then here in
the denominator, I'm going to the square
root of this business, the square root of 8
cosine squared theta. So the square root
of that is going to be 2 square roots of 2. The square root of 8
is 2 square roots of 2. So let me write this. So let me make it
clear what I'm doing. So this right over here is going
to be the square root of this, which is 2 square roots of 2. That's the square root of 8. And the square root
of cosine squared theta is going to
be cosine theta. Now, you might be
saying, hey, wait, if I take the square root
of something squared, then wouldn't that just be the
absolute value of cosine theta? In order to take away
the absolute value, I'd have to assume that
cosine theta is positive. But we can make that assumption
that cosine theta is positive, because if we look right here at
this part of our substitution, if we wanted to solve for theta,
you'd divide both sides by 2, and you'd get x of 2 is
equal to sine of theta. Or we could say that theta is
equal to arcsin of x over 2. Now the arcsin function, as
it is traditionally defined, will return theta that is
between negative pi over 2 and pi over 2. And in that range,
cosine of theta is always going to be positive. So we don't have to
write the absolute value. We know cosine
theta is positive. So now we can start to simplify. Cosine theta cancels
out with cosine theta. This 2 cancels out with this 2. We can bring this square
root of 2 outside. And so we are left with
pi over this square root of 2 times the indefinite
integral of just d theta. And this is just going
to be equal to pi over the square root of 2
times theta plus c. And we're almost done. We just have to rewrite
this in terms of x. And we already know that theta
is equal to arcsin of x over 2. So we can say that this
indefinite integral, or the antiderivative
of this expression, is going to be pi over
the square root of 2 times arcsin of x over 2 plus c. And we're done. Some people like a square
root of 2 in the denominator. If you want to remove
it, you can multiply this by square root of 2
over square root of 2, and that will simplify it. But right now, I'll just
leave the denominator in irrational form. And this right over here
is our antiderivative.