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Current time:0:00Total duration:5:45

Trig and u substitution together (part 2)

Video transcript

in the last video in order to in order to evaluate this indefinite integral we first made the substitution that X is equal to 3 sine theta and then this got us to an integral of this form then we were able to break up these sines and cosines and use a little bit of our trig identities to get it into the form where we could do use substitution we did another substitution where we set that U is equal to cosine of theta and then finally we were able to get it into a form using that second round of substitution and this time it was u substitution we were able to get in to form that we could actually take the antiderivative and we got the final answer here in terms of U but now we got to go and undo everything we have to undo the substitutions so the first substitution that or the last substitution we had done we're not going to go in reverse order was that u was equal to cosine theta so you might just want to you might just want to substitute U with cosine theta here but then we're going to have everything in terms of cosine theta which still doesn't get us to X so the ideal is if I can somehow Express u in terms of X so let's think how we can do it we know that U is equal to cosine theta we know the relationship between X and theta is right over here X is equal to 3 sine theta so let's write that over here so we know that X right over here we know that X X is X is equal to 3 3 sine theta so if we could somehow write cosine let me rewrite this a little bit differently or we could also say that X over 3 is equal to sine theta I just divided both sides by 3 so if we could somehow reexpress this in terms of sine theta then we can replace all the sine Thetas with x over 3 s and we are done so how can we do that and actually show you two techniques for doing it so the first one is to make the realization okay U is equal to cosine of theta if I want to write this in terms of sine of theta I can just say that this is equal to straight-up this is the most fundamental trigonometric identity this is cosine theta is 1 minus the square or the square root of 1 minus sine squared theta and we see sine of theta is equal to x over 3 so this is the square root of 1 minus 1 minus x over 3 1 minus x over 3 squared so u this is U in terms of X so everywhere we see a you up here we can replace it with this expression and we are essentially done we would have written this in terms of X now there's another technique you might sometimes see in a calculus class where someone says okay we know that U is equal to cosine theta we know this relationship how can we express u in terms of X and we'll just say let's draw a right triangle they'll draw a right triangle like this I'll draw a right triangle and they'll say okay look sine of theta is x over three so if we say that this is theta right over here sine of theta is the same thing as opposite over hypotenuse opposite over hypotenuse is equal to x over three so let's say that this is X and then this right over here is three then the sine of theta will be x over three so we look at that first substitution right over here but in order to figure out what U is in terms of X we need to figure out what cosine of theta is well cosine is adjacent over hypotenuse so we have to figure out what this adjacent side is well we could just use the Pythagorean theorem for that Pythagorean theorem would tell us that this is going to be the square root of the hypotenuse squared which is nine minus the other side squared minus x squared so from this we fully kind of solved the right triangle in terms of X we can realize that cosine of theta is going to be equal to the adjacent side square root of nine minus x squared over the hypotenuse over three which is the same thing as 1/3 times the square root of 9 minus x squared which is the same thing if we square one-third and put it into the radical so we're essentially going to take the what the square one-third is the same thing as the square root of 1/9 so this is going to we can rewrite this as the square root of 1/9 times 9 minus x squared essentially just brought the 1/3 into the radical now it's 1/9 and so now this is going to be the same thing as the square root of 1 1 minus x squared 1 minus x squared over 9 which is exactly this thing right over here x squared over 9 is the same thing as x over 3 squared so either way you get the same result I find using the trig identity right over here to express cosine of theta in terms of sine theta and then just do this substitution to be a little bit more straightforward but now we can just substitute into the original thing so either of these I can write it as either way this thing right over here this is the same thing as 1 minus x squared over 9 to the 1/2 power that's what u is equal to and we everywhere we see you we just substitute it with this thing so our final answer in terms of X is going to be equal to 243 times U to the fifth this to the fifth power over 5 this to the fifth power is 1 minus x squared over 9 it was to the 1/2 but if we raise it to the fifth power stop I'm going to be to the 5 1/2 power over 5 over 5 minus this to the this to the 3rd power 1 minus x squared over 9 to the three-halves racing this to the 3rd power that's this right over here over 3 and then all of that plus C and we're done it's messy but using first trig substitution then use substitution or trig substitution then rearranging using a couple of our techniques for manipulating these powers of trig functions we're able to get into a form of record use u substitution and then we were able to unwind all the substitutions and actually get evaluate the indefinite integral