Integration by Substitution

"Integration by Substitution" (also called "u-substitution") is a method to find an  integral , but only when i...

"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 our integral in this form:

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

Like in this example:

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

When our integral is set up like that, we can do this substitution:

Then we can integrate f(u), and finish by putting g(x) back as 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 = 3∫cos(x²) 2x dx

(We can pull constant multipliers outside the integration, see Rules of Integration.)

Then go ahead as before:

3∫cos(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 = ½∫2x/(x²+1) dx

Then we have:

Then integrate:

½∫1/u du = ½ ln(u) + C

Now put u=x²+1 back again:

½ ln(x²+1) + C

And how about this one:

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

When we can put an integral in this form:

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

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