 This lesson is on Udu substitution or an integral in the form of U, some sort of function, and the derivative of that function. Let's study the answers that you had in your lesson where you take derivatives using the chain rule. The first one said f of x is equal to e to the 2x squared. So if we want to find the derivative of that, I'm going to use a little different notation and say df is equal to the derivative of e to the 2x squared, which is e to the 2x squared times the derivative of 2x squared, which is 4x, and I'm going to put my dx on that side. So what do we have here? Well, if we make u equal to 2x squared, we find that the du is equal to 4x dx. And take note of that as you go through these, that we have that u du form sitting right in the problem when we do the derivatives. Let's do a second one. g of x is equal to sine of 4x minus 4. If we do the derivative of g, we take the derivative of sine, which is cosine, 4x, and then we take the derivative of 4x, which is 4. And then we multiply that times dx. And you remember the derivative of negative 4, which is a constant, is 0. So if we look at the u du form here, we see that u this time is the 4x. So u would be 4x, and du would be the 4dx. Again, u du form. If we take the derivative of y is equal to x to the 4th minus 2 to the 11th power, we get dy is equal to 11 times x to the 4th minus 2 to the 10th times the derivative of x to the 4th minus 2, which is 4x cubed. And of course, we'll put a dx there. Our u this time is what's in the middle here, which is x to the 4th minus 2. Our du is 4x cubed. Again, we have a constant, which will be 0 times dx. Again, u du form. And this time it was a power rule, so inside that power rule was our u. On the next one, we have y is equal to sine squared 3x. The derivative is dy is equal to 2 sine 3x cosine 3x times 3. In this case, our u is sine 3x. Our du becomes cosine 3x times 3 dx. And if we go backwards, we will get the sine squared 3x back. Again, u du form. Our last example is z of x is ln of 1 plus x squared plus 3. And if we take the dz of this, we get 1 over 1 plus x squared times the derivative of 1 plus x squared, which is 2x dx. And in this case, our u is equal to 1 plus x squared. And our du is equal to 2x dx, u du form. So as we do these, we have to think this u du form. Let's do some examples of integrals. Evaluate the following integral, the sine of 2x. Well, that does look kind of familiar. If we make u equal to 2x, then our du becomes 2 dx. So as we look at our integral, we see we have the sine of u. But our du is 2 dx, which means we need a 2 inside there to make the du. Well, if I introduce a 2 there, I need to introduce a 1 half out front. And that compensates for having the 2 in there originally. So we've put this into the form sine u du. Now we can evaluate it. The anti-derivative of sine u is negative cosine u plus c. And of course we have our multiplier of 1 half. Replacing the u with what it is, we have the final answer of negative 1 half cosine of 2x plus c. So we look for that u du form, replace everything. The only thing we can compensate for are the numbers. And in this case it was a 2. Putting the 2 in for our du right there, making this all of the du, compensating with the 1 half to make sure that everything looks the same as when we started out. Let's try another one. Evaluate the integral of the square root of 3x dx. In this case, we see that we have something to the 1 half power. So our u is going to be 3x. Then our du is going to be 3 dx. So our integral will now look like the integral of u to the 1 half power. Instead of dx being plain, we have 3 dx. So we have to introduce the 3 here, compensate outside with the 1 third. And so now we have u to the 1 half, 3 dx is our du. So we have 1 third u to the 1 half du. We take the antiderivative or integral of that. We have u to the 3 halves times 2 thirds plus c. That all becomes 2 ninths. Then the u is 3x to the 3 halves plus c. Again, u was 3x because we had this as a power. We're thinking chain rule when we do these things. You know, what is inside that power, that power being the 1 half power. Let's try another one. Evaluate the integral secant 5x tan 5x dx. Well, that looks like secant x tan x. And we know the antiderivative of secant x tan x is secant x. So we can just say that u in this case is 5x. That makes our du 5 dx. So our integral now becomes secant u tan u. And remember, we had substitute of 5 in so that we can have that du, which means we compensate with the 1 fifth out front. And so we bring the 1 fifth down and then make our du there. So our integral now becomes 1 fifth integral secant u tan u du. And the antiderivative of secant tan is secant. So now we have 1 fifth secant u plus c. Substituting for u, we have 1 fifth secant of 5x plus c. Hopefully these are getting a little simpler for you to see as we go along. Now this one looks a little bit more complicated. But again, we look for an antiderivative that we know, an integral that we know in each of these. And if you look at this one, you know the antiderivative of secant squared. So most likely you would be that square root of x because that's a complicated piece of our secant squared x. So yes it is. And as it works out, you'll see u is equal to x to the 1 half. And then the du becomes 1 half x to the negative 1 half dx. And we see all of that is sitting here except for that 1 half. The square root of x, that's x to the negative 1 half, the dx is there, we need a 1 half on this side. Well, if we put a 1 half in with a dx, then we have to put a 2 to compensate on the outside. So now what do we have? We have 2 integral secant squared du. And the antiderivative of secant squared is tangent. So this becomes 2 tangent u plus c. And again, we substitute for u and get 2 tangent of the square root of x plus c. A little bit more complicated. But again, look for the antiderivative that you can take and then look to see where you can place that u. Here is another interesting one. The integral of 1 over the square root of 3x minus 5 dx. Well, that is equal to the integral of 3x minus 5 to the negative 1 half dx. Putting it in that form, you see that u is not a very difficult thing to find. u is 3x minus 5, which makes du equal to 3 dx. Most of the time, our du's have been very, very simple. There's just been numbers which we can put in and compensate for. So now we have 1 third integral of u to the negative 1 half du. If we take that integral, we get 1 third u to the 1 half times 2 plus c. And make a final substitution in cleaning up. It becomes 2 thirds. You do square root of u, put it back in the same form, 3x minus 5 and plus c. Our final example on this is evaluate the integral 2x cubed over x to the fourth minus 1. Now, what would be that u du form here? Well, we see we have x to the fourth down here. And its derivative would be an x cubed. So the most likely suspect here is making u equal to the x to the fourth minus 1. We take that whole denominator. We do not take just one piece of it. So let's make u equal to x to the fourth minus 1 and du very nicely becomes 4x cubed dx. So if we begin to substitute in, the u is in the denominator. For our du, we have 4x cubed. Well, this is 2x cubed. So we have to multiply the numerator by 2, compensate with a 1 half out front. So now we have that 1 half sitting there. So now we have 4x cubed dx, which is our du. So we have du over u. That evaluates to ln of the absolute value of u plus c. Substituting in, we get 1 half ln of x to the fourth minus 1 plus c. Hope this helps you doing u du substitution. They can become very tricky if you do not look for some basics of what has to be substituted for u and du. The last thing I want to leave you with is, when you have your integral ready to take that integral or anti-derivative, it has to be totally in u and du forms. They cannot be any other variable in there. You're only compensating with numbers. This concludes our lesson for today.