Integration by Parts (DP IB Analysis & Approaches (AA)): Revision Note
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Integration by parts
What is integration by parts?
Integration by parts is generally used to integrate the product of two functions
However reverse chain rule and/or substitution should be considered first
e.g.
can be solved using reverse chain rule or the substitution
Integration by parts is essentially ‘reverse product rule’
Though note that whilst every product can be differentiated, not every product can be integrated (analytically)
What is the formula for integration by parts?
This is given in the formula booklet alongside its alternative form
How do I use integration by parts?
For a given integral, parts of the integrand are assigned to be
and
(rather than
and
)
Generally, the function that becomes simpler when differentiated should be assigned to
There are various stages of integrating using this method
However only one overall constant of integration (“+c”) is required
Put this in at the last stage of working
If it is a definite integral then “+c” is not required at all
STEP 1
Name the integral to save rewriting it later (is often used)
e.g.
STEP 2
Assignand
Differentiateto find
and integrate
to find
e.g.
STEP 3
Apply the integration by parts formulae.g.
STEP 4
Work out the second integral,
Now include a “+c” (unless definite integration)e.g.
STEP 5
Simplify the answer if possible or apply the limits for definite integratione.g.
In trickier problems other rules of differentiation and integration may be needed
Chain, product or quotient rule
Reverse chain rule, substitution
Can integration by parts be used when there is only a single function?
Some single functions (non-products) are awkward to integrate directly
e.g.
,
,
,
These can be integrated using parts however
Rewrite as the product ‘
’ and choose
and
1 is easy to integrate and the functions above have standard derivatives listed in the formula booklet
Examiner Tips and Tricks
If or one of the inverse trig functions are one of the functions involved in the product, then these should be assigned to "
" when applying parts.
These functions are (relatively) easy to differentiate (to find ), but are awkward to integrate.
Worked Example
a) Find .

b) Show that .

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Repeated integration by parts
When will I have to repeat integration by parts?
In some problems, applying integration by parts still leaves the second integral as a product of two functions of
Integration by parts will need to be applied again to the second integral
This occurs when one of the functions takes more than one differentiation to become simple enough to make the second integral straightforward
This often involves integrands of the form
How do I apply integration by parts more than once?
STEP 1
Name the integral to save rewriting it later (is often used)
STEP 2
Assignand
. Find
and
STEP 3
Apply the integration by parts formula
STEP 4
Repeat STEPS 2 and 3 for the second integral
STEP 5
Work out the second integral and include a “+c” if necessary
STEP 6
Simplify the answer or apply limits
What if neither function ever becomes simpler when differentiating?
It is possible that integration by parts will end up in a seemingly endless loop
Consider the product
The derivative of
is
No matter how many times a function involving
is differentiated, it will still involve
The derivative of
is
would then have derivative
, and so on
No matter how many times a function involving
or
is differentiated, it will still involve
or
This loop can be 'trapped' by spotting when the second integral becomes identical to (or a multiple of) the original integral
Naming the original integral (
) at the start helps
then appears twice in integration by parts
e.g.
whereis parts of the original integral that have already been integrated
It is then straightforward to rearrange and solve the problem
e.g.
Don't forget to include a constant of integration ('+c') at the end
Examiner Tips and Tricks
'+c' just represents 'any constant', so it doesn't matter that we've divided by 2 between the final two steps above.
I.e. you don't need to worry about changing '+c' to '+
c' in the final step there
But you do need the
in front of
for the answer to be correct!
Worked Example
a) Find .

b) Find .

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