Integration by Parts (College Board AP® Calculus BC): Revision Note

Integration by parts

What is integration by parts?

• Integration by parts is a technique that can be used to integrate a product of two functions

  • Not all products can be integrated using this technique

• The formula is $\int u·\frac{dv}{dx}dx=u·v-\int \frac{du}{dx}·vdx$

• Integration by parts is the reverse of the product rule for differentiation

  • Product rule

    • $\frac{d}{dx}\left(u·v\right)=u·\frac{dv}{dx}+\frac{du}{dx}·v$

  • Integrate each term with respect to $x$

    • $u·v=\int u·\frac{dv}{dx}dx+\int \frac{du}{dx}·vdx$

  • Rearrange

    • $\int u·\frac{dv}{dx}dx=u·v-\int \frac{du}{dx}·vdx$

How do I use integration by parts?

• STEP 1
  Label one function $u$ and one function $\frac{dv}{dx}$

  • E.g. for $\int x\cos xdx$ let $u=x$ and $\frac{dv}{dx}=\cos x$

• STEP 2
  Find the derivative of $u$and the antiderivative of $\frac{dv}{dx}$

  • You do not need to include a constant of integration

  • E.g. $\frac{du}{dx}=1$ and $v=\sin x$

• STEP 3
  Substitute into the formula

  • $\int u·\frac{dv}{dx}dx=u·v-\int \frac{du}{dx}·vdx$

  • E.g. $\int x\cos xdx=x·\sin x-\int 1·\sin xdx$

• STEP 4
  Find the antiderivative of $\frac{du}{dx}·v$

  • E.g. $\int 1·\sin xdx=-\cos x$

• STEP 5
  Simplify and include a constant of integration

  • E.g. $\int x\cos xdx=x\sin x-\left(-\cos x\right)+C=x\sin x+\cos x+C$

How do I choose the function to use for u?

• The trick is to choose $u$ such that $\frac{du}{dx}·v$ is a function that can be integrated

• Remember $\frac{dv}{dx}$ needs to be integrated

  • If a function does not have a straightforward antiderivative, then choose this as $u$

• The order of choice for $u$ is the following:

  • Logarithms and inverse trigonometric functions

    • E.g. $u=\ln x$, $u=\arctan x$ or $u=\mathrm{arc}\sin x$

  • Polynomials

    • E.g. $u=x$ or $u=2x+1$ etc

  • Exponentials and trigonometric functions

    • Though it is very rare you would choose these to be $u$

    • E.g. $u=e^{2x}$ or $u=\sin 3x$

How can I use integration by parts with definite integrals?

• You can use integration by parts with definite integrals

• One method is to apply the limits at each stage

  • The formula looks like ${\int }_a^{b}u·\frac{dv}{dx}dx={\left[u·v\right]}_a^b-{\int }_a^b\frac{du}{dx}·vdx$

• Alternatively, you can find the indefinite integral and then apply the limits

How do I find the antiderivatives of logarithmic and inverse trigonometric functions?

• You can use integration by parts to find antiderivatives of logarithmic and inverse trigonometric functions

• Let $u$ equal the function and set $\frac{dv}{dx}$ equal to 1

  • e.g. for $\int \ln xdx$ use $u=\ln x$ and $\frac{dv}{dx}=1$

• Follow the steps for integration by parts to get the antiderivatives:

  • $\int \ln xdx=x\ln x-x+C$

  • $\int \arctan xdx=x\arctan x-\frac{1}{2}\ln \left(x^2+1\right)+C$

  • $\int \mathrm{arc}\sin xdx=x\mathrm{arc}\sin x+\sqrt{1-x^2}+C$

Can I use integration by parts twice?

• You can use integration by parts twice to find the antiderivative of functions such as $x^2{e}^x$, $x^2\sin x$ or $x^2\cos x$

• Using integration by parts once on these functions will result in an integral of the form $x{e}^x$, $x\cos x$ or $x\sin x$

• Use integration by parts again to complete finding the antiderivative

What is a DI table?

• A DI table is a way to set out your work when using integration by parts multiple times

• Identify the expression to differentiate and write this in a column labeled D

  • Continuously differentiate this until you get 0

• Identify the expression to integrate and write this in a column labeled I

  • Continuously integrate this

• To solve the original indefinite integral:

  • Multiply the first term in the D column by the second column in the I column

    • It should give a diagonal pattern

  • Multiply the second term in the D column by the third column in the I column and subtract it from the overall expression

    • You should alternate between adding and subtracting products

  • When you get to the row with zero, you can stop and just add a constant

    • Technically, the formula tells you to integrate the zero with the other term in that row

    • However, the antiderivative of zero is a constant

• For example, consider $\int x^2\sin xdx$

  • Using the table, you get $x^2·\left(-\cos x\right)-2x·\left(-\sin x\right)+2·\cos x+C$

  • This simplifies to $-x^2\cos x+2x\sin x+2\cos x+C$

Will I have to use integration by parts more than twice?

• You will not have to use integration by parts more than twice in an exam question

• However the expression given after using integration by parts twice might contain the original integral

  • e.g. after one use $\int e^x\cos xdx=e^x\sin x-\int e^x\sin xdx$

  • e.g. after two uses $\int e^x\cos xdx=e^x\sin x+e^x\cos x-\int e^x\cos xdx$

• This formula can then be rearranged to make the original integral the subject

  • e.g. rearranging gives $\int e^x\cos xdx=\frac{1}{2}\left(e^x\sin x+e^x\cos x\right)$

• Remember to include a constant of integration

  • e.g. $\int e^x\cos xdx=\frac{1}{2}\left(e^x\sin x+e^x\cos x\right)+C$

Can I still use a DI table?

• You can still use a DI table

• In these cases, you will not reach a zero in the D row

  • Instead, stop when you reach a row that is similar to the starting row

    • Each term in this row should be a multiple of the starting term in that column

    • E.g. $\sin 3x$ and $-9\sin 3x$ in the D column and $e^{2x}$ and $\frac{1}{4}{e}^{2x}$ in the I column

• Follow the same process:

  • Differentiate the term in the D column

  • Integrate the term in the I column

  • Multiply diagonally

  • Alternate between adding and subtracting

• When you get to the row that is the same row as the first row

  • you multiply both terms in that row

  • integrate the product

  • and add or subtract depending on where the pattern is up to

• This then gives a term that is the same as the starting term

  • You can form and solve equations like the example above

• For example, consider $\int e^x\cos xdx$

  D	I
  $e^x$	$\cos x$
  $e^x$	$\sin x$
  $e^x$	$-\cos x$

  • Using the table, you get $\int e^x\cos xdx=e^x\sin x-e^x·\left(-\cos x\right)+\int e^x·\left(-\cos x\right)dx$

  • This simplifies to $\int e^x\cos xdx=e^x\sin x+e^x\cos x-\int e^x\cos xdx$

  • And can be solved as above to get $\int e^x\cos xdx=\frac{1}{2}\left(e^x\sin x+e^x\cos x\right)$

  • Then just remember a constant of integration $\int e^x\cos xdx=\frac{1}{2}\left(e^x\sin x+e^x\cos x\right)+C$