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let's say I have the indefinite integral 1 over the square root of 3 minus 2x squared and of course I have a DX there so right when I look at that there's no obvious I guess traditional method of taking this antiderivative I don't have the derivative of this sitting someplace else in the integral so I can't do traditional u substitution but what I can do is I could I could say well you know this almost looks like some trig identities that I'm familiar with so maybe I can substitute with trig functions so let's see if I can find a trig identity that looks similar to this well our most basic trigonometric identity this comes from the unit circle definition is that the sine squared of theta plus the cosine squared of theta is equal to 1 and then if we subtract cosine squared of theta from both sides we get or if we subtract sine squared of theta from both sides we could do either we could get cosine squared of theta is equal to 1 minus sine squared of theta we could do either way but this all of a sudden this thing right here starts to look a little bit like this maybe I can do a little bit of algebraic manipulation to make this look a lot like that so the first thing I would like to have a 1 here at least that's how my brain works so let's factor out a 3 out of this denominator so this is the same thing as the integral of 1 over the square root of let me factor out a 3 out of this expression 3 times 1 minus 2/3 x squared all I did I did nothing fancy here I just factored the 3 out of this expression that's all I did but the neat thing now is this expression looks a lot like that it's Direction in fact if I substitute if I say that this thing right here this 2/3 x squared if I set it equal to sine squared theta I will be able to use this identity so let's do that let's set 2/3 x squared let's set that equal to sine squared of theta sine squared of theta so if we take the square root of both sides of this equation I get the square root of two over the square root of three times X is equal to the sine of theta if I want to solve for X what do I get and what we're gonna have to solve for both X and for theta so let's do it both ways first let's solve for theta we could get if we solve for theta you get that theta is equal to the arc sine or the inverse sine of square root of 2 over square root of 3x that's if you solve for theta now if you solve for x you just multiply both sides of this equation times the inverse of this and you get X is equal to divide both sides of the equation by this or multiply by the inverse is equal to the square root of 3 over the square root of 2 times the sine of theta and we're going to substitute this with sine squared of theta beta we can't leave this DX out there we have to take the integral with respect to D theta so what's DX with respect to D theta so the derivative of X with respect to theta is equal to square root of 3 over square root of 2 derivative of this with respect to theta is just cosine of theta and if we want to write this in terms of DX we could just write that DX is equal to square root of 3 over the square root of 2 cosine of theta D theta now we're ready to substitute so we can rewrite this expression up here we can rewrite it I'll do it in this reddish color I was using that let me do it in the blue color we can rewrite this expression up here now it's an indefinite integral of DX is on the numerator right I could instead of writing this 1 times DX I could have just written the DX up here that could be a Det DX and just like that you're just multiplying it times DX so what's DX DX is this business I'll do it in yellow DX is this right here so it's the square root of 3 over the square root of 2 cosine theta D theta that's what DX was now the denominator the denominator my equation I have the square root of the square root of 3 times now it's 1 minus now I said 2/3 x squared is equal to sine squared of theta 1 minus sine squared of theta now how can I simplify this well what's 1 minus sine squared of theta that's cosine squared of theta so this thing right here is cosine squared of theta so my indefinite integral becomes the square root of 3 over the square root of 2 cosine theta d-theta all of that over the square root of square root of 3 times the cosine squared of theta that just became cosine squared of theta so let's just take the square root of this bottom part so this is going to be equal to I'll do an arbitrary change of colors square root of 3 over the square root of 2 cosine of theta d-theta all of that over what's the square root of this is equal to the square root of 3 times the square root of cosine squared so times cosine of theta now this simplifies things a good bit I have a cosine of theta divided by cosine of theta those cancel out just get 1 and then I have a square root of 3 up here divided by square root of 3 so those two guys are going to cancel out so my integral simplifies nicely to 1 over square root of 2 D theta or even better I could write this this is just a constant term I could take it out of my integral I could eat it equals 1 over the square root of 2 times my integral of just D theta and this is super easy this is equal to 1 over the square root of 2 times theta plus C plus some constant and you could say that the integral of this is theta plus C and then you multiply the constant times this but it's still going to be some arbitrary constant I think you know how to take the antiderivative of this but are we done well no we want to we want to know our indefinite in terms of X so now we have to reverse substitute so what is Theta we figure that out here theta is equal to arcsin of square root of 2 over square root of 3x so our original indefinite integral which was all of this silliness up here now that I reversed substitute for theta or put X back in there it's 1 over the square root of 2 times theta theta is just this is just arc sine arc sine of square root of 2 over square root of 3x and then I have this constant out here plus C so this right here is the antiderivative the antiderivative of 1 over the square root of 3 minus 2x squared so hopefully you found that helpful I'm going to do a couple of more videos where we go through a bunch of these examples just so you get familiar with them