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let's see if we can take the indefinite integral of sine to the fourth or sine of X to the fourth this sine of X to the fourth DX and like always pause the video and see if you can work through it on your own so if this was a if we had an odd exponent up here whether it was a sine or cosine then the technique I would use if so if this was sine to the third of X I would separate one of the sine of X's out so I would rewrite it as sine squared X times sine of X and then I would convert this using the Pythagorean identity and then when I distribute the sine of X I would be able to use u substitution and we've done that in previous example videos and you could have done this if this cosine to the third of X as well or to the fifth or to the seventh if you had an odd exponent but here we have an even exponent so what do we do and so the technique we will use and I guess you could call it a trick the technique or trick to use is once again you want to just algebraically manipulate this so that you can use integration techniques that we are familiar with and in this case we would want to use the double angle identity so the double angle identity tells us that sine squared of X is equal to is equal to 1/2 times 1 minus cosine of 2x cosine of 2x so how can I apply this over here well our original integral is just the same thing as this is going to be the same thing as integral of sine of x squared all of that squared DX now I can make this substitution so this this this is the same thing as this which is of course the same thing as this by the double angle identity so I can rewrite it as I can rewrite it as the integral of 1/2 times 1 minus cosine of 2x and then all of that squared all of that squared DX and what is that going to be equal to that's going to be equal to well I see 1/2 squared is 1/4 so I can take that out so we get 1/4 times the integral of so I'm just going to square all of this business 1 squared which is 1 minus 2 cosine of 2x plus cosine squared of 2x DX DX now fairly straightforward to take to take the to take the indefinite integral or to take the antiderivative of these two pieces but what do I do here well once again I've got an even exponent let's apply the double angle identity for cosine so we know that cosine squared of 2x is going to be equal to 1/2 times 1 plus cosine of double this angle so cosine of 4x so once again just make the substitution so this is going to be equal to this is going to be equal to actually oh let me just write it this way 1/4 and I'll maybe I'll switch well I'll keep it in this color 1/4 times 1 minus 2 cosine 2 X plus plus plus 1/2 now let's distribute the 1/2 plus 1/2 plus 1/2 cosine of 4x DX cosine of 4x DX let's see I could take this not that one I could take this 1/2 and add it to this one and that's going to get me 3 halves so add those together I'm going to get three halves and so let me rewrite this as I'm in the homestretch really so this is going to be equal to 1/4 times the integral of 3 halves minus 2 cosine of 2x and then let's see the derivative of 4x is 4 so it'd be great if I have 4 out here so let me write this let me rewrite 1/2 as 4 over 8 I'm just really I'm just multiplying and dividing by 4 it's one way to think about it so I could say plus plus 1/8 times 4 cosine of 4x and I'm just doing this so that it's well good and there you can do u substitution but we've had a lot of practice with this a that have function I have it's derivative I can just integrate with respect to the Forex right over there so as you could say it's the reverse chain rule which is really what u substitution is all about and now I'm ready to integrate so this is going to be equal to I think we deserve a little bit of a drumroll one-fourth times three halves X three halves X see derivative of 2x is sitting right over here two so this is going to be minus sine of 2x we can verify this is to see derivative of this is going to be two cosine of 2x yep we have it right over there plus plus 1/8 times well sine of 4x derivative of sine of 4x is going to be 4 cosine of 4x which is exactly what we have there and then homestretch we just write the plus C plus sub constant this is an indefinite integral and we're all done this wasn't the simplest of problems but it also wasn't too bad and it's strangely satisfying to get it done