6.2 Substitution in the Indefinite Integral

Note To understand this section, you should be familiar with antiderivatives, as discussed in the preceding tutorial.

The technique of substitution or change of variables is based on the chain rule for derivatives. In the following example, we evaluate the integral

5x(2x²+1)^-3 dx

Take u to be an expression that is being raised to a power.

u = 2x²+1.

We now follow a fairly mechanical step-by-step procedure:

Step 1: Calculate the derivative of u, and then solve for "dx." We usually put this calculation in a little box to separate it from the calculation of the integral.	u = 2x2+1 du dx = 4x dx = 1 4x du
Step 2: Substitute the expression for u in the original integral, and also substitute for dx.	5x(2x2+1)-3 dx = 5x(u)-3 1 4x du
Step 3: Eliminate the variable x, leaving an integral in u only. Often, as happens here, all the x's will cancel leaving an expression in u only. (Sometimes, they do not -- see below ).	5x(u)-3 1 4x du = 5(u)-3 1 4 du
Step 4: Simplify the integrand. Here, we can slip the constant 5/4 outside the integral sign.	5(u)-3 1 4 du = 5 4 u-3 du
Step 5: Evaluate the simplified integral. Important: Do not substitute back for u until after this step.	5 4 u-3 du = 5 4 u-2 -2 + C = - 5u-2 8 + C
Final Step: Substitute back for u to obtain the result.	5x(2x2+1)-3dx = - 5(2x2+1)-2 8 + C

To evaluate 4x(3x²+3)^1/2 dx, the substitution that results in the simplest integral is u = ?

	x			3x2			3x2+3
	4x(3x2+3)1/2			(3x2+3)1/2			4x

Using the correct subsitution above, the expression for dx in terms of du is ?

	dx = 6u dx			dx = 6x du			dx = 1 6x du
	dx = 1 6 x du			dx = 6 x du			dx = 6x du

Making the correct substitutions above transforms the integral 4x(3x²+3)^1/2 dx to which of following?

	2 3	u1/2 du			4x u1/2 dx			4x u1/2 du
	4u(u1/2) du

The next step is:

	substitute for u			substitute for "du"
	go to the movies			evaluate the integral

The final answer is

	4 9 (3x2+3)3/2 dx			(3x2+3)3/2 + C			2 9 u3/2 + C
	2 3 (3x2+3)1/2 + C			4 9 (3x2+3)3/2 + C			(3x2+3)3/2 dx

Here is one for you to try on your own. Use proper graphing calculator format to input your answers (spaces are ignored). Here are some examples of correctly formatted expressions involving logarithms and exponentials.

Consider the integral

2x-1 (2x2-2x+5)3

dx.

u =
du/dx =
dx =

After substituting, and simplyfying (canceling the x's) the integral becomes:

The final answer is:

	2x-1 (2x2-2x+5)3	dx	=

Sometimes, rather than an expression raised to a power, we have a number (such as e) raised to an expression. In such cases, the following advice is often useful:

Take u to be an expression that appears in the exponent.

Note To review integrals of exponential functions, press the "prev" button on the sidebar to go to the tutorial on antiderivatives.

For the integral 3x²e^-x3+4 dx, a suitable substitution would be u = ?

	3x2e-x3+4			3x2			-x3+4
	x			e-x3+4

Using the above substitution, we find 3x²e^-x3+4 dx = ?

	3x2e-x3+4 + C			x3e-x4/4+4x + C			- x3e-x3+4 + C
	-e-x3+4 + C

Q This is all very well, but what do I do if the x's do not cancel?
A Let us go through one such example and see...

The substitution u = 2x-1 in the integral 3x(2x-1)^1/2 dx yields:

	3xu1/2 du			3 4 u1/2 du			3x 2 u1/2 du
	3 2 u1/2 du			3x u1/2 du			3 2 xu1/2 du

Now, to eliminate any remaining x's, we so the following:

Go back to the equation that gives u as a function of x, and solve it for x.

Here,

u = 2x-1,

x = (u+1)/2.

Substituting this into the integral and completing the calculation yields:

	3x(2x-1)1/2	dx	=

You now have several options

• Try some of the questions in the true/false quiz (warning: it covers the whole of Chapter 6) by pressing the button on the sidebar.
• Try some of the on-line review exercises (press the "Review Exercises" button on the side. Again, these questions cover the whole chapter, but you should now be able to answer I1 through I14.)
• Try some of the exercises from Section 6.2 in Applied Calculus or Section or Section 13.2 of Finite Mathematics and Applied Calculus.