Integration techniques | U substitution
Ok, we now have the basis of integration! As I said in the previous post, I am now going to cover the techniques of integration. I will make a post for each method, because it’s important that you understand each of them before you can move on. In this lesson I teach you how to perform u-substitution, the most used technique in integration.
U-substitution consists in substituting part of, or the whole, function with a letter, usually u (from this “u-substitution”) to make semplifications that will help a lot in the process of integration. Once you let u be something, you take the derivative of that something and write as in this example:
And then you usually find dx:
Let’s take the integral
We know that the integral of sin(x) is -cos(x). If you don’t know what I am talking about, I suggest you check this out: https://addjustabitofpi.wordpress.com/2019/10/03/fundamental-integrals/.
So this is what we are going to do: let’s let u = 7x and take the derivative;
So we can substitute 7x with u and dx with 1/7 du;
We don’t add the “+C” yet because we still need to undo the u-substitution to get back to the x. If you remember, u = 7x, so
That’s it! Easy, right? As you can see, the u-substitution was used to get something we already know. Let’s take another integral:
We know that we need to get rid of the x. When we perform a u-substitution, we don’t want any x to be left; everything has to be in the “u world”. Let’s let u=x, u=sin(x^2) and u=x^2 to see which one will actually help us.
This wasn’t very useful, as we have the same integral, just in the u world.
Too bad… we don’t want any x!
The x’s canceled out! That’s great! No x left. Then remember to undo the substitution.
Let’s try something just slightly more difficult:
Don’t panic! Look at the function, did you realise that tan(x) is sin(x)/cos(x)? This is a huge step!
It would be great if we could get rid of the cos(x) or sin(x), so that we would be left with sin(x). But wait, we can! If we let u = cos(x), du will be -sin(x)dx and so du=-1/sin(x) dx. At this point the sines cancel out and we are left with 1/u! And we know that the integral of 1/u with respect to u is… ln(u).
If we had let u = sin(x), we would have got du=cos(x)dx, dx=1/cos(x) du and so
Nothing cancels out = bad.
U-sub can be performed more than once, but we need to change the letter, so we can use v for the second substitution, w for the third… it’s up to you.
First, we let u = 5x so that we get that dx = du/5. No x’s! Then we let v = sin(x) so that dx = du/cos(x) (it is the same as 1/cos(x) du) and they cancel out nicely.
And now it’s your time! Give these integrals a try and leave the answer in the comments!
I really hope this explanation was clear and that you have understood this very useful method of integration. If you have any questions, doubts, don’t hesitate to leave a comment below! I will be happy to help you.
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