a lecture on integration by parts

The Method of Integration by Parts

This method is used to find indefinite integrals that look like the result of a product ruleMathworldPlanetmath.

  • 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 f(x)𝑑x:

    1. (a)

      Find functions U and V such that UV=f(x). There are many possible choices.

    2. (b)

      U should be a function easy to derive and such that the derivative of U is easier, less complicated than U itself. For example good choices for U are U=x,x2,x3,ex,lnx.

    3. (c)

      V should be a function which is easy to integrate, such that we can find V𝑑x easily and the integral is less complicated than V itself. Good choices for V are V=ex,sinx,cosx. The functions x,x2,x3,lnx are bad choices.

    4. (d)

      Once the functions U and V are chosen, find U=ddxU and V=V𝑑x.

    5. (e)

      Plug in the formula.

    6. (f)

      Solve the new integral UV𝑑x, which if the choices of U and V were good, should be easy.

    7. (g)

      If the new integral is hard, the choices of U and V might be wrong. So repeat the choice.

Again, the method is best explained through examples:

Example 0.1.

Find xex𝑑x. This is a product of two functions and there is no compositionMathworldPlanetmath visible, so we will be using integration by parts. We need two functions U and V such that UV=xex. A good choice for U is x because U=(x)=1 and the derivative is simpler than x itself. A good choice for V=ex because it is easy to integrate, ex𝑑x=ex (and UV=xex). Therefore we apply the formula:

Example 0.2.



We choose U=x and V=cos(x). Then U=1 and V=V𝑑x=sin(x). Apply the formula:

Example 0.3.

Find lnxdx. Although this is not the product of two functions, lnx is a typical example of a function to be integrated by parts. Here is how, let U=lnx and V=1 so that U=1/x and V=x. Look what happens when we use the formula:

Example 0.4.

In some cases we have to use the method of integration by parts twice to get an answer. For example find x2sinxdx. We put U=x2 and V=sinx, and so U=2x and V=sinxdx=-cosx. Thus:


and above, we have seen that in order to find xcosxdx we use integration by parts. Therefore the final anwser is:

Example 0.5.

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 exsinxdx. Let I=exsinxdx. We start with integration by parts, taking




Hmmm…the integral excosxdx looks like the one we started with…we use integration by parts to solve this one. Take U=ex and V=cosx so U=ex and V=sinx. Thus:




Remember that we want to find I, so if we solve for I above, we obtain:


and so I=1/2(-excosx+exsinx)+C.

Title a lecture on integration by parts
Canonical name ALectureOnIntegrationByParts
Date of creation 2013-03-22 15:38:33
Last modified on 2013-03-22 15:38:33
Owner alozano (2414)
Last modified by alozano (2414)
Numerical id 6
Author alozano (2414)
Entry type Feature
Classification msc 26A36
Related topic ALectureOnIntegrationBySubstitution
Related topic ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution
Related topic ALectureOnThePartialFractionDecompositionMethod
Related topic ExampleOfIntegrationByPartsInvolvingAlgebraicManipulation