# What is Integration by Substitution?

Oct 1, 2014

Integration by substitution - also known as the "change-of-variable rule" - is a technique used to find integrals of some slightly trickier functions than standard integrals. It is useful for working with functions that fall into the class of some function multiplied by its derivative.

Say we wish to find the integral

${\int}_{1}^{3} \ln \frac{x}{x} \mathrm{dx}$
We know that $\ln \frac{x}{x} = \ln \left(x\right) \cdot \frac{1}{x}$
and we also know that the derivative of $\ln \left(x\right)$ is $\frac{1}{x}$
What we can do now is change our variable from x to u. We let u equal the undifferentiated function - in this case, $\ln \left(x\right)$.

$\text{Let } u = \ln \left(x\right)$
$\implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$

We can substitute both of these new functions into the original equation.

${\int}_{1}^{3} \ln \frac{x}{x} \mathrm{dx}$
$= {\int}_{1}^{3} \ln \left(x\right) \cdot \frac{1}{x} \mathrm{dx}$
$= {\int}_{1}^{3} u \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

We're not quite there yet! When we change the variable, we also need to change the terminals of the integral.

$\text{When } x = 1 , \ln \left(x\right) = 0$
$\text{When } x = 3 , \ln \left(x\right) = \ln \left(3\right)$
Note that I leave $\ln \left(3\right)$ in exact form.

The next step requires a little bit of "false multiplication". We all know that if we multiply a fraction by its denominator, the denominator disappears - for example:
$\frac{2}{3} \cdot 3 = 2$
$\frac{6}{13} \cdot 13 = 6$
$\frac{a}{b} \cdot b = a$
The same is true with $\frac{\mathrm{du}}{\mathrm{dx}} \cdot \mathrm{dx}$. When (and ONLY when) we have changed our terminals, we can substitute in the new terminals and do this bit of multiplication:

${\int}_{1}^{3} u \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

$= {\int}_{0}^{\ln} \left(3\right) u \cdot \mathrm{du}$

This is now an integral we can work with. The answer will be correct only if you have correctly changed the terminals.

The process is very similar for finding antiderivatives (or indefinite integrals). The only difference is that we have no terminals to work with, so instead of changing the terminals in the middle, we substitute the original variable at the end. Let's say we have our original integral without terminals:

$\text{Let } u = \ln \left(x\right)$
$\implies \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$
$\implies \int \ln \frac{x}{x} \mathrm{dx}$
$= \int u \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$
$= \int u \cdot \mathrm{du}$
$= \frac{{u}^{2}}{2} + c , c \in \mathbb{R}$
Now we can substitute $u = \ln \left(x\right)$ back into the equation:
$\int \ln \frac{x}{x} \mathrm{dx}$
$= {\left(\ln \left(x\right)\right)}^{2} / 2 + c , c \in \mathbb{R}$.