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Current time:0:00Total duration:4:34

AP.CALC:

FUN‑6 (EU)

, FUN‑6.D (LO)

, FUN‑6.D.1 (EK)

- [Instructor] What we're
going to do in this video is get some more practice
identifying when to use u-substitution and
picking an appropriate u. 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 one 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 one over x dx. Now it becomes clear. We have some function, natural log of x, being raised to the 10th power, but we also have its
derivative right over here, one 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
picked natural log of x is because I see something, I see its exact derivative here, something close to its
derivative, in this case its exact derivative. And so then I could say, du dx, du dx is equal to one over x, which means that du is equal to one over x dx. And so here you have it. This right over here is du, 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 du. 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, to actually evaluate
this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something, let's do something interesting here. Let's say the integral of tangent 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 can rewrite tangent in
terms of sine and cosine. So we can 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 the 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, a 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 one over cosine of x times negative sine of x dx. Now does it jump out at
you what our u might be? Well I have a cosine of
x in the denominator, and I have its derivative, so what if I made u equal to cosine of x? u is equal to cosine of x, and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx. And just like that I have my du 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 one over u, one over u du. Which is a much easier
integral to evaluate, and then once you evaluate this, you back-substitute cosine of x for u.

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