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

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

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

Calculus
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