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Main content
Current time:0:00Total duration:4:34
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.D (LO)
,
FUN‑6.D.1 (EK)

Video transcript

what we're going to do in this video is get some more practice identifying when to use u-substitution and picking and appropriate you so let's say we have the indefinite integral of natural log of X to the to the 10th power all of that over X DX does u substitution apply and if so how would we make that substitution well the key for u substitution is to see do I have some function and its derivative and you might immediately recognize that the derivative of natural log of X is equal to 1 over X to make it a little bit clearer I could write this as the integral of natural log of X to the 10th power times 1 over X DX now it's clear we have some function natural log of X being raised to the tenth power but we also have its derivative right over here 1 over X so we could make the substitution we could say that U is equal to the natural log of X and the reason why I pick natural log of X is because I see something I see its exact derivative here or something close to its derivative in this case it's its exact derivative and so then I could say D u DX D u DX is equal to 1 over X which means that D U is equal to 1 over X DX and so here you have it this right over here is d u and then this right over here is our u and so this nicely simplifies to the integral of U to the 10th power u to the 10th power D U and so you would evaluate what this is find the antiderivative here and then you would back substitute the natural log of X for u and to actually evaluate this indefinite integral let's do another one let's say that we have the integral of let's do it let's do something let's do something interesting here let's say the integral of and X DX does u-substitution apply here and at first you say well I just have a tangent of X where is its derivative but one interesting thing to do is well we could rewrite tangent in terms of sine and cosine so we could write this as the integral of sine of X over cosine of X DX and now you might say well where does u-substitution apply here well there's a couple of ways to think about it you could say the derivative of sine of X is cosine of X but you're now dividing by the derivative as opposed to multiplying by it but more interesting you could say the derivative of cosine of X is negative sine of X we don't have a negative sine of X but we can do a little bit of Engineering we can multiply by negative one twice so we could say the negative of the negative sine of X and I stuck one of them you could say negative ones outside of the integral which comes straight from our integration properties this is equivalent I can put a negative on the outside and negative on the inside so that this is the derivative of cosine of X and so now this is interesting in fact let me rewrite this this is going to be equal to negative the negative integral of 1 over cosine of x times negative sine of X DX now does it jump out at you what are you might be well I have a cosine of X in the denominator and I have its derivative so what if I made you equal to cosine of X U is equal to cosine of X and then D u DX would be equal to negative sine of X or I could say that D U is equal to negative sine of X DX and just like that I have my D u here and this of course is my U and so my whole thing has now simplified to it's equal to the negative indefinite integral of 1 over u 1 over u D U which is a much easy you're integral to evaluate and then once you evaluate this you back substitute cosine of X for you
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