_-substitution: defining _ (more examples) | AP Calculus AB | Khan Academy

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 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? 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. 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 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. 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 something interesting here. Let's say the integral of tan(X) DX. Does u-substitution apply here?

And at first, you might 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 interestingly, 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 one’s outside of the integral, which comes straight from our integration properties. This is equivalent; I can put a negative on the outside and 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 1 over cosine of X times negative sine of X DX.

Now, does it jump out at you? What you 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 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 easier integral to evaluate. And then, once you evaluate this, you back substitute cosine of X for U.