Home Math Teaching About Me

3. Week 2: Integrals and Substitutions

≪ 2. Week 1: Riemann Sums and Summation Notation | Table of Contents | 4. Week 3: Integration by Parts ≫

Over the course of the past few lectures, you have hopefully seen the dogmatic Fundamental Theorem of Calculus, which states (loosely) that the operations of differentiation and integration are each other’s inverses.

When we learned to differentiate things, we started with the simplest functions — polynomials, exponentials, logarithms, trig functions — and then learned rules such as the product rule and the chain rule to handle more exotic and complicated combinations of them. One may therefore be led to believe that these differentiation rules have integral counterparts. This is to some extent true, and the first example of this is the chain rule.

The substitution method consists of several steps:

1. Identify a substitution. Usually it’ll be the stuff inside one of the “simple functions” we already know how to integrate, like \((\star)^5\) or \(\frac{1}{\star}\) or \(\cos(\star)\).
2. Perform the substitution, rewriting the integrand and the bounds of integration (if applicable) in terms of the new variable.
3. Compute the integral, which is hopefully easier at this point. If it’s an indefinite integral, rewrite the answer in terms of the original variables.

Let’s now walk through several illustrative examples to more clearly depict the above pipeline.

1. I know how to integrate \(\cos(\star)\), so I should pick a substitution \(u = 2x\) to make the integrand more closely resemble a “simple function”.
2. If \(u = 2x\), then \(\frac{du}{dx} = 2\), or \(dx = \frac{du}{2}\). Substituting these in to our integral yields \[ \int \cos(2x) dx = \int \cos(u) \frac{du}{2} = \frac{1}{2}\int \cos(u) du. \]
3. We can now recognise that the integral is \[ \frac{1}{2}\sin u + C = \frac{1}{2}\sin(2u) + C. \]

This was a pretty straightforward example that can sometimes just be “eyeballed”.

1. I know how to integrate \((\star)^4\), so I’ll pick the substitution \(u = e^x + 2\) to make the integrand resemble \(u^4\).
2. If \(u = e^x + 2\), then \(\frac{du}{dx} = e^x\), or \(dx = \frac{du}{e^x}\). Substituting these in yields \[ \int _{0}^{\ln 3} e^x \left( e^x + 2 \right)^4 dx = \int _{x=0}^{x=\ln 3}e ^x u^4 \frac{du}{e^x} = \int _{u = 3}^{u = 5} u^4 du. \] Note the limits of integration! The bounds \(0\) and \(\ln 3\) refer to values of \(x\), not to values of \(u\).
3. We can now complete the integral \[ \int _{3}^{5} u^4 du = \left. \frac{1}{5}u^5\right\rvert_{3} ^{5} = \frac{1}{5} \left( 5^5 - 3^5 \right). \]

Sometimes, as in this example, the substitution is not immediately obvious. For instance, if I pick \(u = 1 + \ln x\) (very reasonable), using \(\frac{du}{dx} = \frac{1}{x}\) or \(dx = x\ du \) yields

\[ \int \frac{u \left( 2 - \ln x \right)}{x}x\ du = \int u \left( 2 - \ln x \right) du. \]

Although \(2 - \ln x\) didn’t get cancelled out when we rewrote \(dx\) in terms of \(u\), we can still express \(2 - \ln x = 3 - u \)! Thus our integral becomes

\[ \begin{align*} \int u \left( 3 - u \right) du &= \int \left( 3u - u^2 \right) du \\ &= \frac{3u^2}{2} - \frac{u^3}{3} + C \\ &= \frac{3(1 + \ln x)^2}{2} - \frac{(1 + \ln x)^3}{3} + C. \end{align*} \]