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let's attempt to take the antiderivative or the indefinite integral of X to the third time's the square root of 9 minus x squared DX and you might attempt to do something like u substitution but you'll find that you're not getting very far and we have a big clue here we have something of the form so 9 minus x squared u can be viewed that can be viewed as the same thing as 3 squared minus x squared and anytime you have the form a squared minus x squared it might be useful to make the substitution that X is equal to a sine of theta now why would that be well then a squared minus x squared would become would become a squared minus a squared sine squared theta which is the same thing as a squared times 1 minus sine squared theta I think you see where this is going this is a squared times cosine squared of theta which might be a useful simplification so let's do it over here in this case our a is 3 so let's make the substitution that X is going to be equal to a sine theta or 3 times sine of theta and then when we and then we're going to also have to figure out what DX is equal to so if you take the derivative and we will get DX we could have DX D theta is equal to 3 cosine theta or we could write that DX if we wanted wanted to write it in differential form we could write that DX is equal to 3 times cosine theta D theta this is just the derivative of this with respect to theta and we're ready to substitute back our original expression now becomes it now becomes I'll write in that original green it now becomes 3 sine theta to the third power which is the same thing as 27 sine actually let me color code it just so you know what parts I'm doing so this part right over here X to the third is now going to become 27 sine to the third power of theta or sine theta to the third power and then all of this business 9 minus X this is going to be the square root of nine minus x squared so minus nine sine squared theta and then DX let me do this in a new color DX right over here is going to be equal to that's not a new color DX DX is going to be equal to all of this business so times three times three cosine theta D theta now we let's see if we can simplify this business a little bit the square root let me do this over the side the square root this thing right over here can be written as 9 times 1 minus sine squared theta which is equal to the square root of 9 times cosine squared theta and we can assume we can assume that cosine squared or the cosine theta is positive as we did in the last video and so this is going to be equal to this is going to be equal to 3 cosine theta3 3 cosine theta in orange so this right over here is 3 cosine theta and so what does this simplify to we have a 27 times 3 is 81 times 3 times 3 is going to be 243 so this is going to be 243 I'll put it out front times the integral of so you're going to sine cubed theta sine cubed theta and then you're going to have cosine theta times cosine theta we could say cosine squared theta cosine squared theta that's this term right over here and this term right over there and of course D D theta I think I've taken care of everything so it might not look like I've simplified it a lot because hey look this still doesn't seem like a trivially easy problem to solve but we are getting closer now this turns into just a classic u substitution problem and it's not obvious just yet it's actually a little bit of a lair of a technique to figure out first how do you do u sub tuition right over here and the key when you when you have powers of trig functions especially when you have one of them is an odd power what you want to do is separate one of those odd powers out so you can kind of construct a u-substitution problem so let's do that so this is going to be equal to 243 times the integral of sine cubed theta I can rewrite as sine of theta actually let me write it this way as sine squared theta sine squared theta cosine squared theta cosine squared theta and then I still have one sine theta-1 a sine theta right over here D theta and D theta and what I'm trying to do is turn this expression to something where I can do use substitution and as you could imagine maybe where the D U has to deal with sine theta D theta and it would if I can get my U being equal to cosine theta then my D U is going to be negative sine theta D theta so let me see if I can do that so if I if I say that sine squared theta is the same thing as 1 minus cosine squared theta then this whole thing becomes 243 times the integral of 1 minus cosine squared theta times cosine squared theta times cosine squared theta times sine theta D theta times sine theta D theta and I want to be very clear what I did over here so we use some trig substitution to get to this point right over here and at this point I took one of the sine Thetas out I separated it right over here and then I converted this to an expression in terms of cosine theta now the whole reason why I did this is right over here I have a function of cosine theta and then I have something that's pretty close to the derivative of cosine theta right over here so this is now ripe for you substitution so let's do u substitution if I have some function of something and then I have its derivative maybe you should be equal to that something so let me set u as being equal to cosine of theta cosine of theta then D U is going to be equal to negative sine of theta D theta well I have a sine theta D theta I can multiply that is a negative as long as I put a negative out here I'm multiplying by a negative twice I'm not changing the value so notice now this right over here is d u and this right over here is a function of U so let's write it that way all of this business is going to be equal to negative 243 times the integral of 1 1 minus u squared u squared times u squared times u squared and then this right over here is just our D U so this is just D U now this is pretty straightforward we can just multiply we can just multiply our u out and this will become this is going to become negative 243 so this is all equal negative 240 3 times the indefinite integral of U squared u squared minus U to the fourth u squared minus U to the fourth I'm just distributing the U squared D u D u now this is pretty straightforward to take the antiderivative of this is negative 243 times the antiderivative of U squared is U to the third or 3 antiderivative of U to the fourth is U to the fifth over 5 and of course we're going to have a plus C out here and just so that we can get rid of this negative we can swap this we can distribute the negative sign so this one becomes negative that one becomes positive and we get and we get all of this business as being equal to as being equal to 243 times U to the fifth over five minus U to the third over 3 plus C and you might say wow finally we are done but we aren't done we have everything in terms of U while our original integral was in terms of X so in the next video we're going to unwind all the substitution we're going to try to take this expression right over here and write it in terms of X so we're going to go from u to theta 2x because we've done two rounds of substitutions you