U Substitution

U-substitution is named for the symbolic technique of computing certain antiderivatives by abstracting away a complicated portion of the integrand and renaming it by a single letter, $u$. When done correctly, this allows one to replace an originally intimidating integral with a much simpler one, compute the new simpler antiderivative, and then use the result to answer the original question with only minimal extra work.

Undoing the Chain Rule

The mechanics of $u$-substitution will make little sense until you appreciate the reason for its existence: it is the antidifferentiation analog of the chain rule from differential calculus! A quick refresher: the chain rule $(f(g(x)))^\prime = f^\prime(g(x))g^\prime(x)$ shows how to compute the derivative of $f(g(x))$ in terms of the derivatives of $f$ and $g$. Taking the indefinite integral of both sides of this expression (and using the fundamental theorem of calculus to note that the integral of $(f\circ g)^\prime$ is just $f\circ g$ itself) we see

$$f\circ g +C =\int f^\prime(g(x))g^\prime(x)dx$$

Is best to concentrate on a few concrete examples. Try to express each of these below as $(f(g(x)))^\prime$ for some $f$ and $g$, and then take the anti-derivative!

$$\int 2\cos(2x),dx$$

$$\int 2x\cos(x^2),dx$$

$$\int 2x\tan(x)+x^2 \sec^2(x),dx$$

The Making of a Memorable Notation

The reason that the reverse of the chain rule is called substitution is that we simplify function composition by substituting a single letter for the ‘inside’ function in a composition. That is, in the expression $\int \cos(x^2)2xdx$ we single out the innermost $x^2$ in the composition $\sin(x^2)$ and rename it $u$. Differentiating $u=x^2$ we see that $u^\prime = \frac{du}{dx}=2x$, and so $du=2xdx$. With these two pieces, we can substitute all of the $x$-expressions in the original integrand for $u$ expressions, converting $\int\cos(x^2)2xdx$ into $\int\cos(u)du$. This second, substituted integral is much easier - we can see immediately that the answer must be $\sin(u)+C$! And then simply recalling that $u=x^2$, we have arrived at the final answer $\sin(x^2)+C$, which is readily confirmed via differentation.

To summarize this process as a sequence of steps we:

  • Step 1: Choose a $u$, and compute $du$.
  • Step 2: Replace all $x$’s in the integrand with $u$’s
  • Step 3: Find the antiderivative of this new (hopefully simpler) integrand.
  • Step 4: Finally, use the definition of $u$ to convert everything back to $x$’s.

Finding a Useful ‘U’

There is no one ‘right’ way to choose some portion of an integrand to call $u$, but choosing a helpful $u$ involves satisfying several constraints:

  • The $u$ should be some expression of $x$’s inside a composition, as we are undoing the chain rule

  • The $u$ should be such that when it is abstracted away by one letter, the integral becomes simpler

  • The $u$ needs to be chosen so that $du$ either shows up in the integrand already, or can be made to do so after some algebraic manipulation.

Here’s a collection of examples to try:

$$\int\frac{1}{\sqrt{4x-3}}dx$$

$$\int\frac{x}{\sqrt{1-4x^2}}dx$$

$$\int e^{5x}dx$$

$$\int x^5\sqrt{1+x^2}dx$$

$$\int\tan(x)dx$$

Some tricky ones:

$$\int z\sqrt{z+1}dz$$

$$\int \frac{x+1}{x(x+2)}dx$$

$$\frac{x}{1+x^4}dx$$

$$\int x^2\sqrt{2+x}dx$$

Substitution for Definite Integrals

Recall that a definite integral (giving the area under a curve) is found by choosing an antiderivative of the integrand, and taking the difference of its values at the endpoints: $\int_a^b f(x)dx=F(b)-F(a)$. One way to do this is just the follow the entire process for calculating an antiderivative above, and then evaluate the endpoints (after switching back to $x$) as normal.

But since we are just looking for an area, this may seem like overkill: after all we are not interested in the antiderivative itself, we are just using it as a tool in our calculation. Can we perhaps avoid the conversion back to $x$ by instead converting the entire definite integral over to the new variable $u$?

If $u$ is our new variable, defined as a function $u(x)$ of $x$, we may think of this function as a translation tool from the $x$-axis to the $u$-axis. Thus, if our original bounds were $x=a$ and $x=b$, then the conversion to $u$-bounds is simply $c=u(a)$ and $d=u(b)$!

$$\int_a^b f(u(x))u^\prime(x)dx=\int_{u(a)}^{u(b)}f(u)du$$

Below are some examples showing both the original area on the $x$-axis (blue) and the substituted area on the $u$-axis (green). Even though the two curves have quite different shapes in each case, the area underneath is always exactly the same!