Figuring out u substitutions

By Tutor GuyNo Comments


To determine if a ‘u substitution’ is appropriate for an integral, look for the value of du somewhere in the original integral. You may need to multiply or divide by a constant to get du in the integral.

Example 1: Integrate:

\displaystyle \int 3x^2e^{x^3} \, dx

Here the obvious choice for u \text{ is } x^3. And the derivative of x^3=3x^2, so du is in the integral too. Make the substitutions and solve:

\displaystyle \int 3x^2e^{x^3} \, dx = \int e^u \, du=e^u+C=e^{x^3}+C

Example 2: Integrate:

\displaystyle \int (x+1)(x^2+2x)^3 \, dx

 Do you see that x + 1 is the derivative of x^2+2x if we multiply x+1 by 2? Therefore, let u=x^2+2x, and du=2x+2=2(x+1) \, dx. To get the factor of 2 in the integral, multiply by 2 and multiply on the outside by ^1 \!\! / \! _{2.} The problem then becomes:

\displaystyle \int (x+1)(x^2+2x)^3 \, dx= \frac{1}{2} \int 2(x+1)(x^2+2x)^3 \, dx

\displaystyle = \frac{1}{2} \int (u)^3 \, du= \frac{1}{8} u^4+C= \frac{1}{8}(x^2+2x)^4+C

Example 3: Integrate:

\displaystyle \int \tan x \, dx

How is this a u substitution problem? First, rewrite \tan x:

\displaystyle \int \tan x \, dx= \int \frac{\sin x \, dx}{\cos x}

 Now do you see the u sub? Let u= \cos x \text{ and }du=- \sin x \, dx. Multiply inside and outside the integral by -1 so you can make the following substitution and solve:

\displaystyle \int \tan x \, dx= \int \frac{\sin x \, dx}{\cos x}=- \int \frac{- \sin x \, dx}{\cos x}=- \int \frac{du}{u}

=- \ln |u|+C=- \ln | \cos x|+C

What if you let u= \sin x in this example? Then du= \cos x \, dx. This seems like a reasonable substitution. However, this will put dx in the denominator. And that’s a definite no no! Any time you make a u substitution, the du term must be in the numerator.

Note: There are a couple of instances where a u substitution works and du isn’t in the original integral. You will learn these with a little practice. Here’s one example. Note that the radical term is not the derivative of the x term and the x term is not the derivative of the radical. However, with a little rearranging, a u substitution leads to an expression you can integrate:

\displaystyle \int x \sqrt{x+1} \, dx

\text{Let } u=x+1. \text{ Then } du=dx \text{ and } x=u-1. Substituting gives

\displaystyle \int (u-1) \sqrt{u} \, du= \int u^{3/2}-u^{1/2} \, du= \frac{2}{5}u^{5/2}- \frac{2}{3}u^{3/2}+C

= \dfrac {2}{5}(x+1)^{5/2}- \dfrac{2}{3}(x+1)^{3/2}+C

Blue Taste Theme created by Jabox