Substitution with  x=sin(theta)

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.