U-Substitution

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U-Substituion: Basics :: Definite Integrals :: Use with Functions of U :: Practice Problems :: Quiz

BASICS: (top)

U-substitution is one of the simplest integration techniques that can be used to make integration easier. In its most basic form, u-substitution is used when an integral contains some function and its derivative, that is, for an integral of the form int(f(x)f'(x)dx). The integration is achieved by rewriting the integral in a form that makes it easier to read. Here, let u=f(x). Then du/dx=f'(x), so (with the aid of an engineering hat), we can say du=f'(x)dx and the integral becomes int(udu). This integral can now be easily evaluated; we know int(udu)=(u^2)/2+C. The substitution is then reversed, giving us ((f(x))^2)/2+C.

EXAMPLE 1:

Problem: Evaluate int(sinxcosxdx).

Solution: We see that this integral contains both the function sinx and its derivative, . Therefore, we write u=sinx ; du=cosxdx and the integral becomes int(udu). Integration can then proceed normally: int(udu)=(u^2)/2+C. We now reverse our replacement, substituting sinx for u. Thus, we obtain .
We can check that this answer is correct by differentiating, and, indeed, d/dx((sinx)^2/2+C)=sinxcosx.

DEFINITE INTEGRALS: (top)

When evaluating a definite integral using u-substitution, how does one deal with the limits of integration? There are two possibilities. As before, we have with an integral of the form int,a,b(f(x)f'(x)dx). We will again let u=f(x), giving us du=f'(x)dx. When we make the substitution in the integral, the limits of integration also change, giving us int,u1,u2(udu). Normal integration, as before, gives us int=. But what are these new limits of integration u1 and u2? We can regard them simply as placeholders. When we reverse our replacement, substituting sinx for u, the limits of integration also revert back to the original a and b. This gives us , which can be evaluated in the standard fashion.
Alternatively, after finding int=, we could use the fact that since u=f(x), u1=f(a) and u2=f(b) to evaluate directly without reversing the u-substitution. Using this method, we have .

EXAMPLE 2:

Problem: Evaluate int,.

Solution:

Method 1:

As in Example 1, we see that this integral contains both the function sinx and its derivative, cosx, and make the substitution u=sinx ; du=cosxdx, giving us int,u1,u2(udu). Integrating, reversing the substitution, and evaluating, we get .

Method 2:

As in method 1, we substitute and integrate to find . Since u=f(x), u1=f(0)=sin(0)=0 and u2=f(pi/2)=sin(pi/2)=1. Thus, we have , which is the same as the answer we found using method 1.

INTEGRATION WITH FUNCTIONS OF U: (top)

In our discussion thus far, we have always been left with the integral int(udu), whether in definite or indefinite form. We can also use u-substitution when the integral that remains after we make the substitution is a more complicated function of u, int(f(u)du). This makes u-substitution a much more powerful tool, as it can be used to solve many more problems.

EXAMPLE 3:

Problem: Evaluate .

Solution: We make the same substitution as in examples 1 and 2, u=sinx ; du=cosxdx. This gives the integral , which can be evaluated normally exactly as int,u1,u2(udu) was. In this case, we find .

We have discussed

PRACTICE PROBLEMS: (top | solutions)

Evaluate the following using u-substitution:

1. (solution)

2. (solution)

3. (solution)

QUIZ: (top)

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This page last updated 23 July, 2008 11:38 AM