# 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 $u$.

## Substitution Rule ## Example Integration by Substitution

Example 1. Find the integral $$\int \cos(x^2) 2x dx$$

We know (from above) that it is in the right form to do the substitution: Now integrate: $$\int \cos(u) du = \sin(u) + C$$ And finally put $u=x^2$ back again: $$\sin(x^2) + C$$ So $\displaystyle \int\cos(x^2) 2x dx = \sin(x^2) + C.$$That worked out really nicely! (Well, I knew it would.) But this method only works on some integrals of course, and it may need rearranging: Example 2. Find the integral$$\int \cos(x^2) 6x dx$$Oh no! It is 6x, not 2x like before. Our perfect setup is gone. Never fear! Just rearrange the integral like this:$$\int\cos(x^2) 6x dx = 3\int\cos(x^2) 2x dx$$(We can pull constant multipliers outside the integration, see Rules of Integration.) Then go ahead as before:$$3\int\cos(u) du = 3 \sin(u) + C$$Now put u=x^2 back again:$$3 sin(x^2) + C$$Done! Now let’s try a slightly harder example: Example 3. Find the integral$$\int \frac{x}{x^2+1} dx$$Let me see … the derivative of x^2 +1 is 2x … so how about we rearrange it like this:$$\int \frac{x}{x^2 +1} dx = \frac{1}{2}\int \frac{2x}{x^2 +1} dx$$Then we have: Then integrate:$$\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln(u) + C$$Now put u=x^2 +1 back again:$$\frac{1}{2}\ln(x^2 +1) + C$$And how about this one: Example 4. Find the integral$$\int (x+1)^3 dx$$Let me see … the derivative of x+1 is … well it is simply 1. So we can have this:$$\int (x+1)^3 dx = \int (x+1)^3 \cdot 1 dx$$Then we have: Then integrate:$$\int u^3 du = \frac{1}{4}\cdot u^4 + C$$Now put u=x+1 back again:$$ \frac{(x+1)^4}{4} + C$$We can take that idea further like this: Example 5. Find the integral$$\int (5x+2)^7 dx$$If it was in THIS form we could do it:$$\int (5x+2)^7\cdot 5dx$$So let’s make it so by doing this:$$\frac{1}{5} \int (5x+2)^7\cdot 5dx$$The \frac{1}{5} and 5 cancel out so all is fine. And now we can have u=5x+2 And then integrate:$$\frac{1}{5} \int u^7 du = \frac{1}{5} \cdot \frac{1}{8} u^8 +C$$Now put u=5x+2 back again, and simplify:$$\frac{(5x+2)^8}{40} + C$$Now get some practice, OK? ## In Summary • When we can put an integral in this form: • Then we can make$u=g(x)$and integrate$\displaystyle \int f(u) du$• And finish up by re-inserting$g(x)$where$u$is$g(x)\$. 