Tag Archives: Parts

Integration by Parts 2

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Sometimes when doing an Antidifferentiation by Parts, the resulting integral is simpler than the one you started with but requires another, perhaps several more, antidifferentiations. You can do this but it can get a little complicated keeping track of everything especially with all the minus signs. There is an easier way.

Let’s consider an example: \int{{{x}^{4}}\sin \left( x \right)dx}.

Begin by making a table as shown below. After the headings:

    • In the first column leave the first cell blank and then alternate plus and minus signs down the column.
    • In the second column leave the first cell blank and then enter u and then under it list its successive derivatives.
    • In the third column enter dv in the first cell and then list its successive antiderivatives under it

Tabular 5

The antiderivative is found by multiplying across each row starting with the row with the  first plus sign and adding the products:

\int{{{x}^{4}}\sin \left( x \right)dx}=

-{{x}^{4}}\cos \left( x \right)+4{{x}^{3}}\sin \left( x \right)+12{{x}^{2}}\cos \left( x \right)-24x\sin \left( x \right)-24\cos \left( x \right)+C

Integration by Parts 1

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The antidifferentiation technique known as Integration by Parts or Antidifferentiation by Parts is based on the formula for the Product Rule: d\left( uv \right)=udv+vdu.
Solve this equation for the second term on the right: udv=d\left( uv \right)-vdu.
Integrating this gives the formula \int{udv}=\int{d\left( uv \right)}-\int{vdu}. By the FTC the first term on the right can be simplified giving the formula for Integration by Parts:

\int{udv}=uv-\int{vdu}

The technique is used to find antiderivatives of expressions such as \int{x\sin \left( x \right)dx} in which there is a combination of functions that are usually of different types – here a polynomial and a trig function.

The parts of the integrand must be matched to the parts of \int{udv}. Here we make the substitutions u=x,dv=\sin \left( x \right)dx and from these we compute du=dx\text{ and }v=-\cos \left( x \right). (There is no need for the +C here; it will be included later). Making these substitutions gives

\int{x\sin \left( x \right)dx}=-x\cos \left( x \right)-\int{-\cos \left( x \right)dx}

The integral on the right is simple so we end with

\int{x\sin \left( x \right)dx}=-x\cos \left( x \right)+\sin \left( x \right)+C

As the problems get more difficult the first question students ask is which part should by u and which dv? The rules of thumb are (1) Chose u to be something that gets simpler when differentiated, and (2) chose dv to be something you can antidifferentiated or at least something that does not get more complicated when you antidifferentiate. For example, in the problem above if we were to choose u=\sin \left( x \right) and  dv=xdx the result is

\int{x\sin \left( x \right)dx}=\tfrac{{{x}^{2}}}{2}\sin \left( x \right)-\int{\tfrac{{{x}^{2}}}{2}\cos \left( x \right)dx}

This is correct, but the integral on the right is more complicated than the one we started with. When this happens, go back and start over.

For AP Calculus teachers: Note that Antidifferentiation by Parts is a BC only topic. It is something you can do in AB classes after the AP exam if you have time.