Substitution for Indefinite Integrals

Integration by substitution, or $u$-substitution, is the most common technique of finding an antiderivative.  It allows us to find the antiderivative of a function by reversing the chain rule.  To see how it works, consider the following example.

Let $f(x)=(x^2-2)^8$.  Then  $f'(x)=8(x^2-2)^7(2x)$ by the chain rule.  Remember that we have the "inside part" $(x^2 -2)$, and its derivative $(2x)$.  We can use this knowledge to antidifferentiate, if we have an integrand that has an "inside part" and is multiplied by the derivative of that "inside part".

To see this, consider $\int 8(x^2-2)^7(2x)\,dx$.  We make a substitution to make it clear what to do:  Let $u$ be the inside part, so $u=x^2-2$.  We differentiate $u$, and use the notation $du=2x\,dx$ (instead of $\frac{du}{dx}=2x$; you will see why when we subsitute back into the integrand).  Then $\int 8(x^2-2)^7(2x)\,dx=\int 8u^7\,du$ by directly replacing every element in the first integrand, including the $dx$, by exactly what it becomes via the substution (in blue above).  This integral is now easy to evaluate, giving $u^8+c=(x^2-2)^8+c$.

Notice our integrand above was a composite function, $f(g(x))$ where $g(x)=x^2-2$ was the inside part.  This $u$-substitution process can be stated formally as shown below.  Notice that the first integrand is a composite function multiplied by the derivative of the inside part.

Another example:

$\displaystyle \int \cos(x^3)\cdot 3x^2\,dx\overset{\fbox{$ \,\,u\,=\,x^3,\\ du\,=\,3x^2\,dx$}\\}{=}\int \cos(u)\,du= \sin(u)+c=\sin(x^3)+c$.    Do:  Differentiate to check this answer!

By choosing a suitable function $u$, we can often convert hard integrals into much easier integrals that we know how to evaluate. Unfortunately, there is no magic formula for deciding what $u$ should be — this is a toolkit, not a recipe.  Some examples are in the video that follows.