U-Substitution

### THE METHOD OF U-SUBSTITUTION

The following problems involve the method of u-substitution. It is a method for finding antiderivatives. We will assume knowledge of the following well-known, basic indefinite integral formulas :
1. , where a is a constant
2. , where k is a constant
The method of u-substitution is a method for algebraically simplifying the form of a function so that its antiderivative can be easily recognized. This method is intimately related to the chain rule for differentiation. For example, since the derivative of ex is

,

it follows easily that

.

However, it may not be obvious to some how to integrate

.

Note that the derivative of can be computed using the chain rule and is

.

Thus, it follows easily that

.

This is an illustration of the chain rule "backwards". Now the method of u-substitution will be illustrated on this same example. Begin with

,

and let

u = x2+2x+3 .

Then the derivative of u is

.

Now "pretend" that the differentiation notation is an arithmetic fraction, and multiply both sides of the previous equation by dx getting

or

du = (2x+2) dx .

Make substitutions into the original problem, removing all forms of x , resulting in

= e u + C

= e x2+2x+3 + C .

Of course, it is the same answer that we got before, using the chain rule "backwards". In essence, the method of u-substitution is a way to recognize the antiderivative of a chain rule derivative. Here is another illustraion of u-substitution. Consider

.

Let

u = x3+3x .

Then (Go directly to the du part.)

du = (3x2+3) dx = 3(x2+1) dx ,

so that

(1/3) du = (x2+1) dx .

Make substitutions into the original problem, removing all forms of x , resulting in

.

Most of the following problems are average. A few are challenging. Make careful and precise use of the differential notation dx and du and be careful when arithmetically and algebraically simplifying expressions.

The following problems require u-substitution with a variation. I call this variation a "back substitution". For example, if u = x+1 , then x=u-1 is what I refer to as a "back substitution".