AP®MathsCollege BoardCalculus BCRevision NotesUnit 6: Integration & Accumulation of ChangeMethods of IntegrationIntegration by Parts

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

Dan Finlay

Written by: Dan Finlay

Reviewed by: Mark Curtis

Updated on

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 integral u times fraction numerator d v over denominator d x end fraction d x equals u times v minus integral fraction numerator d u over denominator d x end fraction times v d x

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

    • Product rule

      • fraction numerator d over denominator d x end fraction open parentheses u times v close parentheses equals u times fraction numerator d v over denominator d x end fraction plus fraction numerator d u over denominator d x end fraction times v

    • Integrate each term with respect to x

      • u times v equals integral u times fraction numerator d v over denominator d x end fraction d x plus integral fraction numerator d u over denominator d x end fraction times v d x

    • Rearrange

      • integral u times fraction numerator d v over denominator d x end fraction d x equals u times v minus integral fraction numerator d u over denominator d x end fraction times v d x

How do I use integration by parts?

  • STEP 1
    Label one function u and one function fraction numerator d v over denominator d x end fraction

    • E.g. for integral x cos x d x let u equals x and fraction numerator d v over denominator d x end fraction equals cos x

  • STEP 2
    Find the derivative of uand the antiderivative of fraction numerator d v over denominator d x end fraction

    • You do not need to include a constant of integration

    • E.g. fraction numerator d u over denominator d x end fraction equals 1 and v equals sin x

  • STEP 3
    Substitute into the formula

    • integral u times fraction numerator d v over denominator d x end fraction d x equals u times v minus integral fraction numerator d u over denominator d x end fraction times v d x

    • E.g. integral x cos x d x equals x times sin x minus integral 1 times sin x d x

  • STEP 4
    Find the antiderivative of fraction numerator d u over denominator d x end fraction times v

    • E.g. integral 1 times sin x d x equals negative cos x

  • STEP 5
    Simplify and include a constant of integration

    • E.g. integral x cos x d x equals x sin x minus open parentheses negative cos x close parentheses plus C equals x sin x plus cos x plus C

Examiner Tips and Tricks

In your working, be sure to clearly identify what you are using for u and fraction numerator d v over denominator d x end fraction and clearly show the results for fraction numerator d u over denominator d x end fraction and v.

How do I choose the function to use for u?

  • The trick is to choose u such that fraction numerator d u over denominator d x end fraction times v is a function that can be integrated

  • Remember fraction numerator d v over denominator d x end fraction 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 equals ln x, u equals arctan x or u equals arc sin x

    • Polynomials

      • E.g. u equals x or u equals 2 x plus 1 etc

    • Exponentials and trigonometric functions

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

      • E.g. u equals e to the power of 2 x end exponent or u equals sin 3 x

Examiner Tips and Tricks

You can use the acronym LIPET to help you remember how to select the function for u. LIPET stands for logarithms, inverse trig, polynomials, exponentials and trig.

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 integral subscript a superscript b u times fraction numerator d v over denominator d x end fraction d x equals open square brackets u times v close square brackets subscript a superscript b minus integral subscript a superscript b fraction numerator d u over denominator d x end fraction times v d x

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

Worked Example

Find the indefinite integral integral 5 x e to the power of 3 x end exponent d x.

Answer:

STEP 1
Label one function u and one function fraction numerator d v over denominator d x end fraction

u equals 5 x and fraction numerator d v over denominator d x end fraction equals e to the power of 3 x end exponent

STEP 2
Find the derivative of uand the antiderivative of fraction numerator d v over denominator d x end fraction

table row cell u equals 5 x end cell blank cell fraction numerator d v over denominator d x end fraction equals e to the power of 3 x end exponent end cell row cell fraction numerator d u over denominator d x end fraction equals 5 end cell blank cell v equals 1 third e to the power of 3 x end exponent end cell end table

STEP 3
Substitute into the formula

integral 5 x e to the power of 3 x end exponent d x equals 5 x times 1 third e to the power of 3 x end exponent minus integral 5 times 1 third e to the power of 3 x end exponent d x

STEP 4
Find the antiderivative of fraction numerator d u over denominator d x end fraction times v

table row cell integral 5 times 1 third e to the power of 3 x end exponent d x end cell equals cell 5 over 3 integral e to the power of 3 x end exponent d x end cell row blank equals cell 5 over 3 times 1 third e to the power of 3 x end exponent end cell row blank equals cell 5 over 9 e to the power of 3 x end exponent end cell end table

STEP 5
Simplify and include a constant of integration

integral 5 x e to the power of 3 x end exponent d x equals 5 over 3 x e to the power of 3 x end exponent minus 5 over 9 e to the power of 3 x end exponent plus C

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 fraction numerator d v over denominator d x end fraction equal to 1

    • e.g. for integral ln x d x use u equals ln x and fraction numerator d v over denominator d x end fraction equals 1

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

    • integral ln x d x equals x ln x minus x plus C

    • integral arctan x d x equals x arctan x minus 1 half ln open parentheses x squared plus 1 close parentheses plus C

    • integral arc sin x d x equals x arc sin x plus square root of 1 minus x squared end root plus C

Worked Example

Find the value of integral subscript 0 superscript 1 arctan x d x.

Answer:

Rewrite this as 1 times arctan x and choose the inverse trig function to differentiate

table row cell u equals arctan x end cell blank cell fraction numerator d v over denominator d x end fraction equals 1 end cell row cell fraction numerator d u over denominator d x end fraction equals fraction numerator 1 over denominator x squared plus 1 end fraction end cell blank cell v equals x end cell end table

Use the integration by parts formula and include the limits

integral subscript 0 superscript 1 arctan x d x equals open square brackets x times arctan x close square brackets subscript 0 superscript 1 minus integral subscript 0 superscript 1 fraction numerator x over denominator x squared plus 1 end fraction d x

You can evaluate the first term

table row cell open square brackets x times arctan x close square brackets subscript 0 superscript 1 end cell equals cell open parentheses arctan 1 close parentheses minus 0 end cell row blank equals cell pi over 4 end cell end table

Evaluate the new integral by noticing that the numerator looks like the derivative of the denominator

table row cell integral subscript 0 superscript 1 fraction numerator x over denominator x squared plus 1 end fraction d x end cell equals cell 1 half integral subscript 0 superscript 1 fraction numerator 2 x over denominator x squared plus 1 end fraction d x end cell row blank equals cell 1 half times open square brackets ln open parentheses x squared plus 1 close parentheses close square brackets subscript 0 superscript 1 end cell row blank equals cell 1 half times open parentheses ln 2 minus ln 1 close parentheses end cell row blank equals cell 1 half ln 2 end cell end table

Combine the two terms

integral subscript 0 superscript 1 arctan x d x equals pi over 4 minus 1 half ln 2

You could have also found the indefinite integral before apply the limits

integral arctan x d x equals x times arctan x minus 1 half ln open parentheses x squared plus 1 close parentheses plus C

table row cell integral subscript 0 superscript 1 arctan x d x end cell equals cell open square brackets x times arctan x minus 1 half ln open parentheses x squared plus 1 close parentheses close square brackets subscript 0 superscript 1 end cell end table

integral subscript 0 superscript 1 arctan x d x equals pi over 4 minus 1 half ln 2

Can I use integration by parts twice?

  • You can use integration by parts twice to find the antiderivative of functions such as x squared e to the power of x, x squared sin x or x squared cos x

  • Using integration by parts once on these functions will result in an integral of the form x e to the power of 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 integral x squared sin x d x

    • Using the table, you get x squared times open parentheses negative cos x close parentheses minus 2 x times open parentheses negative sin x close parentheses plus 2 times cos x plus C

    • This simplifies to negative x squared cos x plus 2 x sin x plus 2 cos x plus C

D

I

x squared

sin x

2 x

negative cos x

2

negative sin x

0

cos x

Worked Example

Find the indefinite integral integral x squared cos open parentheses 3 x close parentheses d x.

Answer:

Use integration by parts with u equals x squared and fraction numerator d v over denominator d x end fraction equals cos open parentheses 3 x close parentheses

table row cell u equals x squared end cell blank cell fraction numerator d v over denominator d x end fraction equals cos open parentheses 3 x close parentheses end cell row cell fraction numerator d u over denominator d x end fraction equals 2 x end cell blank cell v equals 1 third sin open parentheses 3 x close parentheses end cell end table

table row cell integral x squared cos open parentheses 3 x close parentheses d x end cell equals cell x squared times 1 third sin open parentheses 3 x close parentheses minus integral 2 x times 1 third sin open parentheses 3 x close parentheses d x end cell row blank equals cell 1 third x squared sin open parentheses 3 x close parentheses minus 2 over 3 integral x sin open parentheses 3 x close parentheses d x end cell end table

Use integration by parts again on the new integral with u equals x and fraction numerator d v over denominator d x end fraction equals sin open parentheses 3 x close parentheses

table row cell u equals x end cell blank cell fraction numerator d v over denominator d x end fraction equals sin open parentheses 3 x close parentheses end cell row cell fraction numerator d u over denominator d x end fraction equals 1 end cell blank cell v equals negative 1 third cos open parentheses 3 x close parentheses end cell end table

table row cell integral x sin open parentheses 3 x close parentheses d x end cell equals cell x times open parentheses negative 1 third cos open parentheses 3 x close parentheses close parentheses minus integral 1 times open parentheses negative 1 third cos open parentheses 3 x close parentheses close parentheses d x end cell row blank equals cell negative 1 third x cos open parentheses 3 x close parentheses plus 1 third integral cos open parentheses 3 x close parentheses d x end cell row blank equals cell negative 1 third x cos open parentheses 3 x close parentheses plus 1 third times 1 third sin open parentheses 3 x close parentheses end cell row blank equals cell negative 1 third x cos open parentheses 3 x close parentheses plus 1 over 9 sin open parentheses 3 x close parentheses end cell end table

Substitute this back in to the full integral
Remember to multiply by the factor in front of the integral

table row cell integral x squared cos open parentheses 3 x close parentheses d x end cell equals cell 1 third x squared sin open parentheses 3 x close parentheses minus 2 over 3 open parentheses negative 1 third x cos open parentheses 3 x close parentheses plus 1 over 9 sin open parentheses 3 x close parentheses close parentheses end cell end table

Simplify and include a constant of integration

table row cell integral x squared cos open parentheses 3 x close parentheses d x end cell equals cell 1 third x squared sin open parentheses 3 x close parentheses plus 2 over 9 x cos open parentheses 3 x close parentheses minus 2 over 27 sin open parentheses 3 x close parentheses plus C end cell end table

The DI table for this would look like

D

I

x squared

cos 3 x

2 x

1 third sin 3 x

2

negative 1 over 9 cos 3 x

0

negative 1 over 27 sin 3 x

table row cell integral x squared cos open parentheses 3 x close parentheses d x end cell equals cell x squared times open parentheses 1 third sin 3 x close parentheses minus 2 x times open parentheses negative 1 over 9 cos 3 x close parentheses plus 2 times open parentheses negative 1 over 27 sin 3 x close parentheses plus C end cell end table

This simplies to

table row cell integral x squared cos open parentheses 3 x close parentheses d x end cell equals cell 1 third x squared sin open parentheses 3 x close parentheses plus 2 over 9 x cos open parentheses 3 x close parentheses minus 2 over 27 sin open parentheses 3 x close parentheses plus C end cell end table

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 integral e to the power of x cos x d x equals e to the power of x sin x minus integral e to the power of x sin x d x

    • e.g. after two uses integral e to the power of x cos x d x equals e to the power of x sin x plus e to the power of x cos x minus integral e to the power of x cos x d x

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

    • e.g. rearranging gives integral e to the power of x cos x d x equals 1 half open parentheses e to the power of x sin x plus e to the power of x cos x close parentheses

  • Remember to include a constant of integration

    • e.g. integral e to the power of x cos x d x equals 1 half open parentheses e to the power of x sin x plus e to the power of x cos x close parentheses plus C

Examiner Tips and Tricks

If you find rearranging with integrals tricky, then assign a variable to the integral. For example, let I equals integral e to the power of x cos x d x then I equals e to the power of x sin x plus e to the power of x cos x minus I. You should find this easier to rearrange.

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 3 x and negative 9 sin 3 x in the D column and e to the power of 2 x end exponent and 1 fourth e to the power of 2 x end exponent 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 integral e to the power of x cos x d x

    D

    I

    e to the power of x

    cos x

    e to the power of x

    sin x

    e to the power of x

    negative cos x

    • Using the table, you get integral e to the power of x cos x d x equals e to the power of x sin x minus e to the power of x times open parentheses negative cos x close parentheses plus integral e to the power of x times open parentheses negative cos x close parentheses d x

    • This simplifies to integral e to the power of x cos x d x equals e to the power of x sin x plus e to the power of x cos x minus integral e to the power of x cos x d x

    • And can be solved as above to get integral e to the power of x cos x d x equals 1 half open parentheses e to the power of x sin x plus e to the power of x cos x close parentheses

    • Then just remember a constant of integration integral e to the power of x cos x d x equals 1 half open parentheses e to the power of x sin x plus e to the power of x cos x close parentheses plus C

Examiner Tips and Tricks

The DI table is a good way to organize your work. However, make sure you understand the process and how it is just repeated use of the formula. You should always state the formula too.

Worked Example

Find the indefinite integral integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x.

Answer:

Use integration by parts with u equals e to the power of 2 x end exponent and fraction numerator d v over denominator d x end fraction equals sin open parentheses 5 x close parentheses

table row cell u equals e to the power of 2 x end exponent end cell blank cell fraction numerator d v over denominator d x end fraction equals sin open parentheses 5 x close parentheses end cell row cell fraction numerator d u over denominator d x end fraction equals 2 e to the power of 2 x end exponent end cell blank cell v equals negative 1 fifth cos open parentheses 5 x close parentheses end cell end table

table row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell e to the power of 2 x end exponent times open parentheses negative 1 fifth cos open parentheses 5 x close parentheses close parentheses minus integral 2 e to the power of 2 x end exponent times open parentheses negative 1 fifth cos open parentheses 5 x close parentheses close parentheses d x end cell row blank equals cell negative 1 fifth e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 5 integral e to the power of 2 x end exponent cos open parentheses 5 x close parentheses d x end cell end table

Use integration by parts again on the new integral with u equals e to the power of 2 x end exponent and fraction numerator d v over denominator d x end fraction equals cos open parentheses 5 x close parentheses

table row cell u equals e to the power of 2 x end exponent end cell blank cell fraction numerator d v over denominator d x end fraction equals cos open parentheses 5 x close parentheses end cell row cell fraction numerator d u over denominator d x end fraction equals 2 e to the power of 2 x end exponent end cell blank cell v equals 1 fifth sin open parentheses 5 x close parentheses end cell end table

table row cell integral e to the power of 2 x end exponent cos open parentheses 5 x close parentheses d x end cell equals cell e to the power of 2 x end exponent times 1 fifth sin open parentheses 5 x close parentheses minus integral 2 e to the power of 2 x end exponent times 1 fifth sin open parentheses 5 x close parentheses d x end cell row blank equals cell 1 fifth e to the power of 2 x end exponent sin open parentheses 5 x close parentheses minus 2 over 5 integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell end table

Substitute this back in to the full integral
Remember to multiply by the factor in front of the integral

table row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell negative 1 fifth e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 5 open parentheses 1 fifth e to the power of 2 x end exponent sin open parentheses 5 x close parentheses minus 2 over 5 integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x close parentheses end cell row blank equals cell negative 1 fifth e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 25 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses minus 4 over 25 integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell end table

Rearrange to make integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x the subject

table row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x plus 4 over 25 integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell negative 1 fifth e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 25 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses end cell row cell 29 over 25 integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell negative 1 fifth e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 25 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses end cell row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell 25 over 29 open parentheses negative 1 fifth e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 25 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses close parentheses end cell row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell negative 5 over 29 e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 29 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses end cell row blank blank blank end table

Include a constant of integration

table row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell negative 5 over 29 e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 29 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses plus C end cell end table

The DI table for this would look like

D

I

e to the power of 2 x end exponent

sin 5 x

2 e to the power of 2 x end exponent

negative 1 fifth cos 5 x

4 e to the power of 2 x end exponent

negative 1 over 25 sin 5 x

integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x equals e to the power of 2 x end exponent times open parentheses negative 1 fifth cos 5 x close parentheses minus 2 e to the power of 2 x end exponent times open parentheses negative 1 over 25 sin 5 x close parentheses plus integral 4 e to the power of 2 x end exponent times open parentheses negative 1 over 25 sin 5 x close parentheses

You can then rearrange as above to get the answer

table row cell integral e to the power of 2 x end exponent sin open parentheses 5 x close parentheses d x end cell equals cell negative 5 over 29 e to the power of 2 x end exponent cos open parentheses 5 x close parentheses plus 2 over 29 e to the power of 2 x end exponent sin open parentheses 5 x close parentheses plus C end cell end table

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Author
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Dan Finlay

Author: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

Mark Curtis

Reviewer: Mark Curtis

Expertise: Maths Content Creator

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.