a lecture on integration by parts
The Method of Integration by Parts
When to use it: Use this method when the integrand is a product of two functions (and when the method of substitution clearly does not work).
How to use it: The method is based in the following formula:
Suppose we want to solve :
Find functions and such that . There are many possible choices.
should be a function easy to derive and such that the derivative of is easier, less complicated than itself. For example good choices for are .
should be a function which is easy to integrate, such that we can find easily and the integral is less complicated than itself. Good choices for are . The functions are bad choices.
Once the functions and are chosen, find and .
Plug in the formula.
Solve the new integral , which if the choices of and were good, should be easy.
If the new integral is hard, the choices of and might be wrong. So repeat the choice.
Again, the method is best explained through examples:
Find . This is a product of two functions and there is no composition visible, so we will be using integration by parts. We need two functions and such that . A good choice for is because and the derivative is simpler than itself. A good choice for because it is easy to integrate, (and ). Therefore we apply the formula:
We choose and . Then and . Apply the formula:
Find . Although this is not the product of two functions, is a typical example of a function to be integrated by parts. Here is how, let and so that and . Look what happens when we use the formula:
In some cases we have to use the method of integration by parts twice to get an answer. For example find . We put and , and so and . Thus:
and above, we have seen that in order to find we use integration by parts. Therefore the final anwser is:
Finally, in some other cases, after we do integration by parts twice, we get to the same integral we wanted to solve. Although it would seem we are stuck, no no! we will be able to find a solution right away.
Find . Let . We start with integration by parts, taking
Hmmm…the integral looks like the one we started with…we use integration by parts to solve this one. Take and so and . Thus:
Remember that we want to find , so if we solve for above, we obtain:
and so .
|Title||a lecture on integration by parts|
|Date of creation||2013-03-22 15:38:33|
|Last modified on||2013-03-22 15:38:33|
|Last modified by||alozano (2414)|