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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 eight minus 2x squared doesn't it's not as obvious and then it's a squared minus x squared but we could simplify this or I guess we can write in a way that it starts to have this pattern you can rewrite eight minus two let me write it right under it you could write eight minus two x squared as if we factor out a 2 as two times four minus x squared and now this very clearly has a pattern a squared minus x squared you could write this as two times two squared minus x squared so in this case a would be equal to two so let's make that substitution let's make the substitution that X is going to be equal to two two sine theta and DX is going to be equal to two 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 two times two squared minus x squared x squared is two sine theta so x squared is going to be two squared sine theta squared and now we can factor out the 2 squared so we're going to get this is going to be two times two squared times one minus sine squared theta two times two squared well that's just going to be eight times cosine squared 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 we're going to have and I'll take the pie outside of the outside of the integration so we're going to pie times DX DX is to 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 DX right over there is to cosine theta D theta so let me write that to 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 get 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 they're going to be the square root of this which is 2 square roots of 2 that's square root of 8 and the square root of cosine squared theta is going to be cosine theta cosine 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 would have to assume that cosine theta is positive but we can assume we can make that assumption that cosine theta is positive because if we look at this part if we look right here at this part of our substitution if we wanted to solve for theta you divide both sides by 2 and you get X of 2 is equal to sine of theta or we could say that theta is equal to arcsin arc sine of x over 2 now the arc sine function as it is traditionally defined will return a theta that is between PI over or between negative PI over 2 and PI over 2 and in that range cosine of theta is always going to be positive so we can we don't have to write the absolute value we know cosine theta is positive so now we can start to simplify it cosine theta cancels that with cosine theta this 2 cancels out with this 2 we could bring this square root of 2 outside and so we are left with PI over this square root of 2 over the 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 x 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 arc sine of X over 2 so we can say that pi this indefinite integral or the antiderivative of this expression is going to be PI over the square root of 2 times arc sine arc sine of X over 2 plus C and we're done some people don't 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 2 but square root of 2 over square root of 2 and that will simplify it but right now I just leave the denominator in irrational form and this right over here is our antiderivative