Week 2 – Part 2

U-Substitution

Remember the chain rule for differentiation?
\[\frac{d}{dx}F(g(x)) = F'(g(x))g'(x)\]
U-substitution is the chain rule done in reverse. Suppose that \(F' = f\).
\[\int f(g(x))g'(x) dx = F(g(x)) + C\]
Now let's substitute \(u = g(x)\).
\[\int f(g(x))g'(x) dx = F(u) + C\]
But we also know that the following is true.
\[\int f(u) du = F(u) + C\]
Therefore, it should be obvious that
\[\int f(g(x))g'(x) dx = \int f(u) du\]
This means that \(du = g'(x) dx\). That's right, we can treat \(du\) as a differential. So how does this help us integrate? We can pick an \(u\) inside the integrand and find its \(du\). Then we substitute the variable \(x\) out completely, so our integrand is a function of \(u\) and we are integrating with respect to \(u\). Once we finish evaluating the integral, we replace \(u\) with whatever we picked when we started the integration.

Let's try a simple example. Suppose we have the following integral.
\[\int \cos (2x+3) dx\]
Let's pick \(u = 2x+3\). Then \(du = 2 dx\). So \(dx = \frac{1}{2}du\). Making these substitutions
\[\int \cos (2x+3) dx = \int \cos (u) \frac{1}{2}du\]
But \(\frac{1}{2}\) is a constant and may be pulled out of the integral. So
\[\int \cos (u) \frac{1}{2}du = \frac{1}{2} \int \cos (u) du\]
Integrating with regard to \(u\), we have
\[\frac{1}{2} \int \cos (u) du = \frac{1}{2}\sin (u) + C\]
Finally, replace \(u\) with \(2x+3\), which is what we chose earlier.
\[ \frac{1}{2}\sin (u) + C =  \frac{1}{2}\sin (2x+3) + C\]
And we are done. Pretty cool, right?

Next week, we start Chapter 6, Applications of Integration. And I am taking my first midterm this week, as well. :'(