# Integration by Substitution

"Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain Rule") 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=x ^{2}** and its derivative

**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(x^{2}) 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=x ^{2}** back again:

sin(x^{2}) + C

So **∫****cos(x ^{2}) 2x dx = sin(x^{2}) + C**

That worked out really nicely! (Well, I knew it would.)

Let's just run through that again in a step-by-step manner:

^{2}) 2x dx

^{2}

^{2}) 2x dx becomes ∫cos(u) du

^{2}) + C

But this method only works on *some* integrals of course, and it may need rearranging:

### Example: ∫cos(x^{2}) 6x dx

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

Never fear! Just rearrange the integral like this:

∫cos(x^{2}) 6x dx = 3∫cos(x^{2}) 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=x ^{2}** back again:

3 sin(x^{2}) + C

Done!

Now let's try a slightly harder example:

### Example: ∫x/(x^{2}+1) dx

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

∫x/(x^{2}+1) dx = ½∫2x/(x^{2}+1) dx

Then we have:

Then integrate:

½∫1/u du = ½ ln**|**u**|** + C

Now put **u=x ^{2}+1** back again:

½ ln(x^{2}+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:

∫u^{3} du = \frac{u^{4}}{4} + C

Now put **u=x+1** back again:

\frac{(x+1)^{4}}{4} + C

We can take that idea further like this:

### Example: ∫(5x+2)^{7} dx

If it was in THIS form we could do it:

∫(5x+2)^{7} **5** dx

So let's make it so by doing this:

\frac{1}{5} ∫(5x+2)^{7} **5** dx

The \frac{1}{5} and 5 cancel out so all is fine.

And now we can have **u=5x+2**

And then integrate:

\frac{1}{5} ∫u^{7} du = \frac{1}{5} \frac{u^{8}}{8} + C

Now put **u=5x+2** back again, and simplify:

\frac{(5x+2)^{8}}{40} + C

## In Summary

**u=g(x)**and integrate ∫f(u) du

**g(x)**where

**u**is.