u substitutions with definite integrals

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When a definite integral requires a u substitution to solve, be sure to substitute for the limits of integration as well. This way, you don’t need to substitute back in for the original function. Instead, you evaluate the integral using the new (u) limits. Here’s an example to show how this works.


\displaystyle \int_0^{\pi/4} \sec^2 \theta \tan^2 \theta \; d \theta

This is an obvious candidate for a u substitution. (See other posts on this website for more information on when to use u substitutions.)

Let u= \tan \theta. Then du=\sec^2 \theta \; d \theta.

But don’t stop there! Use your expression for u to determine the new limits as well.

\theta =0 \rightarrow u=0; \; \theta= \dfrac{\pi}{4} \rightarrow u=1

So the new integral becomes

\displaystyle \int_0^{\pi/4} \sec^2 \theta \tan^2 \theta \; d \theta =\int_0^1 u^2 \; du= \left. \dfrac{1}{3}u^3 \right |_0^1=\dfrac{1}{3}-0=\dfrac{1}{3}

You have found the solution to the original integral without needing to put the integral back in terms of \theta.

Figuring out u substitutions

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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

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