# Integration by Substitution

"Integration by Substitution" (also called "u-substitution") is a method to find an integral, but only when it can be set up in a special way.

The first and most vital step is to be able to write your integral in this form:

Note that you have g(x) and its derivative g'(x)

It might make more sense as is an example:

Here f=cos, and you have g=x2 and its derivative of 2x
This integral is good to go!

If your integral is set up like that, then you can do this substitution:

Then you can integrate f(u), and finish by putting g(x) back as a replacement for u.

Like this:

### Example: ∫cos(x2) 2x dx

We know (from above) that it is in the right form to do the substitution:

Now integrate:

cos(u) du = sin(u) + C

And finally put u=x2 back again:

sin(x2) + C

So cos(x2) 2x dx = sin(x2) + C worked out really nicely! (Well, I knew it would.)

This method only works on some integrals of course, and it may need rearranging:

### Example: ∫cos(x2) 6x dx

Oh no! It is 6x, not 2x. Our perfect setup is gone.

Never fear! Just rearrange the integral like this:

cos(x2) 6x dx = 3cos(x2) 2x dx

(You are allowed to pull constant multipliers outside the integration, see Rules of Integration).

Then proceed as before:

3cos(u) du = 3 sin(u) + C

Now put u=x2 back again:

3 sin(x2) + C

Done!

Now we are ready for a slightly harder example:

### Example: ∫x/(x2+1) dx

Let me see ... the derivative of x2+1 is 2x ... so how about we rearrange it like this:

x/(x2+1) dx = ½2x/(x2+1) dx

Then we have:

Then integrate:

½1/u du = ½ ln(u) + C

Now put u=x2+1 back again:

½ ln(x2+1) + C

### Example: ∫(x+1)3 dx

Let me see ... the derivative of x+1 is ... well it is simply 1.

So we can have this:

(x+1)3 dx = (x+1)3 · 1 dx

Then we have:

Then integrate:

u3 du = (u4)/4 + C

Now put u=x+1 back again:

(x+1)4 /4 + C

So there you have it.

## In Summary

If you can put an integral in this form:

Then you can make u=g(x) and integrate f(u) du

And finish up by re-inserting g(x) where u is.