# 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:

$u&space;=&space;x^2$ $du&space;=&space;2x\,&space;dx$

And then you usually find dx:

$dx=\frac{1}{2x}\,&space;du$

Let’s take the integral

$\int&space;\sin(7x)\,&space;dx$

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.com/2019/10/03/fundamental-integrals/.

So this is what we are going to do: let’s let u = 7x and take the derivative;

$du=7\,&space;dx$

therefore

$dx=\frac{1}{7}\,&space;du$

So we can substitute 7x with u and dx with 1/7 du;

$\int&space;\sin(u)\frac{1}{7}du=\frac{1}{7}\int&space;\sin(u)du=\frac{1}{7}(-\cos(u))=-\frac{1}{7}\cos(u)$

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

$-\frac{1}{7}\cos(u)=-\frac{1}{7}\cos(7x)+C$

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:

$\int&space;x\sin(x^2)dx$

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.

$u=x$ $du=dx\rightarrow&space;dx=du$ $\int&space;u\sin(u^2)du$

This wasn’t very useful, as we have the same integral, just in the u world.

$u=\sin(x^2)$ $du=2x\cos(x^2)dx$ $dx=\frac{1}{2x\cos(x^2)}du$ $\int&space;xu\frac{1}{2x\cos(x^2)}du=\frac{1}{2}\int\frac{u}{\cos(x^2)}du$

Too bad… we don’t want any x!

$u=x^2$ $du=2x\,&space;dx$ $dx=\frac{1}{2x}du$ $\int&space;x&space;\sin(u)\frac{1}{2x}du=\frac{1}{2}\int&space;\sin(u)du=&space;-\frac{1}{2}\cos(u)&space;=&space;-\frac{1}{2}\cos(x^2)+C$

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:

$\int&space;\tan(x)dx$

Don’t panic! Look at the function, did you realise that tan(x) is sin(x)/cos(x)? This is a huge step!

$\int&space;\frac{\sin(x)}{\cos(x)}dx$

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).

$u=\cos(x)$ $du=-\sin(x)\,&space;dx$ $dx=-\frac{1}{\sin(x)}\,&space;du$ $\int&space;\frac{\sin(x)}{u}(-\frac{1}{\sin(x)})dx=-\int&space;\frac{1}{u}du=-\ln(u)=-\ln(\cos(x))+C$

If we had let u = sin(x), we would have got du=cos(x)dx, dx=1/cos(x) du and so

$\int&space;\frac{u}{\cos(x)}(-\frac{1}{\cos(x)})du$

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.

Example:

$\int&space;\sin^6(5x)\cos(5x)dx$

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.

$u=5x$ $du=5\,&space;dx\rightarrow&space;dx=\frac{du}{5}$ $\int&space;\sin^6(u)\cos(u)du$ $v=\sin(x)$ $dv=\cos(u)du\rightarrow&space;du=\frac{dv}{\cos(u)}$ $\int&space;v^6\cos(u)\frac{dv}{\cos(u)}=\int&space;v^6dv=\frac{1}{7}v^7=\frac{1}{7}\sin^7(u)=\frac{1}{7}\sin^7(5x)+C$

And now it’s your time! Give these integrals a try and leave the answer in the comments!

$\int&space;\left&space;(&space;e^x&space;\right&space;)^7dx$ $\int&space;e^{\cos(x)}\sin(x)dx$ $\int&space;\frac{x^{\pi&space;k}}{x}dx$

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.

Support my site!

I put all of my efforts to make this site as good as possible. If you would like, you can help my site grow with a small donation! I would really appreciate it.

\$1.00