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Current time:0:00Total duration:6:06

Video transcript

let's see if we can take the indefinite integral of cosine of X to the third power and so I encourage you to pause the video and see if you can figure this out on your own so you have given it good given it a go and you might have gotten stuck some of y'all might have been able to figure it out but some of y'all might have gotten stuck you're like okay cosine to the third power well gee if I only had a derivative of cosine here if I had a negative sine of X or a sine of X here maybe I could have used to use substitution but how do I take the antiderivative of cosine of X to the third power and the key here is is to use some basic trigonometric identities so what do i mean by that well we know we know that sine squared X plus cosine squared X is equal to 1 or if we subtract sine squared from both sides we know that cosine squared X is equal to 1 right this way is equal to 1 minus sine squared X so what would happen if cosine to the third power that's cosine squared times cosine so what happens if we were to take what it what if we were to what if we were to take what that cosine squared so let me just rewrite it this is the same thing as cosine of X times cosine squared of X DX what if we were to take this thing right over here let me do that magenta color what if we were to take this right over here and replace it with this I know what you're thinking Sal what's that going to do for me this is it feels like I'm making I'm making this this integral even more convoluted and what I would tell you I would say you this this might seem like it's getting more complicated but as you explore and you play with it you'll see that this is actually makes the integral more solvable so let's try it out so if we do that this is going to be equal to this is going to be equal to the integral the indefinite integral cosine of x times 1 minus sine squared X D X and so what is this going to be equal to this is going to be equal to do this in that in that green color this is going to be equal to the indefinite integral of cosine X I just I'm just going to distribute the cosine of X cosine of X minus minus cosine of X cosine of X sine squared of X sine squared sine squared X and then I can close the parentheses DX and this of course is going to be equal to the integral of cosine of X DX and we know what that's going to be minus the integral switch to one color now of cosine X sine squared X sine squared X DX now this is where it gets interesting this part right over here is pretty straightforward the antiderivative of cosine of X is just sine of X so this right over here is going to be sine of X and I'll worry about the plus C at the end because both of these are going to have a plus C so might as well just put one big plus C at the end so that's sine of X and then what do we have going on over here well you might recognize so I have a function of sine of X I'm taking sine of X and I'm squaring it and then I have sine of X is derivative right over here so this fits this fits the I have some derivative of a function and then I have another and then I have a a I guess you say a function of that function so G of f of X and that's a sign that maybe you substitution is in order or we've seen the pattern we've seen this show multiple times already that you could just say well okay if I have if I have a if I'm if I'm a function of a function and I have that functions derivative then essentially I can just take the antiderivative with respect to this function so this would be equal to this would be equal to say capital G is the antiderivative of lowercase D capital G of f of X plus C now if what I just said didn't make sense then we could do u substitution and go through it a little bit more step-by-step so let's just do that because we want things to make sense that's the whole point of these these videos so we could say U is equal to sine of X and then D U is going to be equal to cosine of X DX and so this part and that part is going to be D U and then this is going to be U squared so this is going to be minus we have the integral of U squared D u well what is this going to be this is going to be we're going to have negative U to the third power over three and then we know what u is that U is equal to sine of X so we have our sine of X here for the first part of the integral so for the first integral so we have the sine of X and then this is going to be minus let me just write it this way minus one-third minus one-third instead of U to the third we know unit use sine of X sine of X to the third power and then now now we can throw that plus C there and we've done and we're done we've just evaluated that indefinite integral and the key to it is to just play around a little bit with trigonometric identities so that you can get the integral to a point that you can use the reverse chain rule or you could use u substitution which is just really another way of expressing the reverse the reverse chain rule