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. :'(