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let's see if we can take the indefinite integral of sine squared X cosine to the third X DX and like always pause the video and see if you can work it through on your own alright so right when you look at it you're like oh wow if this was just a sine of X not a sine squared of X well that's going to be the derivative of the derivative of cosine of X maybe I could abuse u-substitution likewise if this was just a cosine of X not a cosine to the third of X I could have used u substitution I could have said u is equal to sine of X but I can't do that over here and so the general idea is here if you have if one of these has an odd exponent and you see that this one does have an odd x one the cosine has an odd exponent what you do is you try to engineer algebraically engineer this expression side so that you can use u substitution the way that you can do that if you have an odd exponent like this is to separate out one of the cosines X and then use the Pythagorean theorem a and identity with the remaining cosine squared X so what do I mean by that so let me just rewrite this so this can be rewritten as sine squared X times let me write it this way times cosine squared X cosine X oh it is I rewrote something to the third power something to the second power times that thing to the first power DX DX and then this can be rewritten as sine squared X and then I'm going to use the Pythagorean identity to convert this into a sine squared or a really one minus sine squared so that by the Pythagorean identity is the same thing as 1 minus sine squared X and then we have cosine X times cosine X DX DX and now I can distribute this sine squared times 1 minus sine squared and I am left with so all of let me do this in a new color so this business right here so now I'm left with the indefinite integral of so sine squared X times 1 is going to be sine squared X and then sine squared x times negative sine squared X is negative sine to the fourth and then all of that times cosine X all of that times cosine X DX now this is starting to look interesting because I have sine squared X minus sine of the fourth X but I have the derivative of sine sitting out here I have cosine X that's the whole reason why we did this little algebraic manipulation and so you substitution works out quite well now because if we say if we said that let me do another color I'll do purple if we say that U is equal to sine is equal to sine of X then D U then D U is going to be equal to cosine of X DX and that works out quite well because we have the D U right over here and then this would be u squared minus U to the fourth which we know how to take the antiderivative of so we're in the homestretch so we can rewrite this as the indefinite integral instead of sine squared X we're saying sine is the same thing as u so we can rewrite that as you can rewrite that as u squared minus minus U to the fourth times D u times D u and this is pretty straightforward now this is going to be u to the third over 3 minus U to the fifth over five plus c plus plus C and then we just do the reverse substitution and then that gets us to be instead of U we want to put a sine of X there so we're going to get sine to the third X or sine of X to the third over 3 minus sine of X to the fifth power write that five a little bit sine of X to the fifth power over five plus C and we are done