Further Techniques of Integration (DP IB Maths: AA HL)

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Paul

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Integration by Substitution

What is integration by substitution?

Integration by substitution is used when an integrand where reverse chain rule is either not obvious or is not spotted

in the latter case it is like a “back-up” method for reverse chain rule

How do I use integration by substitution?

For instances where the substitution is not obvious it will be given in a question

e.g. Find $\int_{}^{} \cot x d x$ using the substitution $u = \sin x$

Substitutions are usually of the form $u = g (x)$

in some cases $u^{2} = g (x)$ and other variations are more convenient

as these would not be obvious, they would be given in a question

if need be, this can be rearranged to find $x$ in terms of $u$

Integration by substitution then involves rewriting the integral, including “ $d x$ ” in terms of $u$

STEP 1
Name the integral to save rewriting it later
Identify the given substitution $u = g (x)$

STEP 2
Find $\frac{d u}{d x}$ and rearrange into the form $f (u) d u = g (x) d x$ such that (some of) the integral can be rewritten in terms of $u$

STEP 3
If limits are involved, use $u = g (x)$ to change them from $x$ values to $u$ values

STEP 4
Rewrite the integral so everything is in terms of $u$ rather than $x$
This is the step when it may become apparent that $x$ is needed in terms of $u$

STEP 5
Integrate with respect to u and either rewrite in terms of $x$ or apply the limits using their $u$ values

For quotients the substitution usually involves the denominator
It may be necessary to use ‘adjust and compensate’ to deal with any coefficients in the integrand
Although $\frac{d u}{d x}$ can be treated like a fraction it should be appreciated that this is a ‘shortcut’ and the maths behind it is beyond the scope of the IB course

Exam Tip

If a substitution is not given in a question, it is usually because it is obvious
- If you can't see anything obvious, or you find that your choice of substitution doesn't reduce the integrand to something easy to integrate, consider that it may not be a substitution question

Worked example

Use the substitution $u = (1 + 2 x)$ to evaluate $\int_{0}^{1} x {(1 + 2 x)}^{7} d x$ .

5-9-2-ib-hl-aa-only-we1-soltn

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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. $\int_{}^{} 2 x \cos (x^{2}) d x$ can be solved using reverse chain rule or the substitution $u = x^{2}$
- Integration by parts is essentially ‘reverse product rule’
  - whilst every product can be differentiated, not every product can be integrated (analytically)

What is the formula for integration by parts?

$\int_{}^{} u \frac{d v}{d x} d x = u v - \int_{}^{} v \frac{d u}{d x} d x$
This is given in the formula booklet alongside its alternative form $\int_{}^{} u d v = u v - \int_{}^{} v d u$

How do I use integration by parts?

For a given integral $u$ and $\frac{d v}{d x}$ (rather than $u$ and $v$ ) are assigned functions of $x$
Generally, the function that becomes simpler when differentiated should be assigned to $u$
There are various stages of integrating in this method

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 if it doesn’t have one already!
This saves having to rewrite it several times – I is often used for this purpose.
e.g. $I = \int_{}^{} x \sin x d x$

STEP 2
Assign $u$ and $\frac{d v}{d x}$ .
Differentiate $u$ to find $\frac{d u}{d x}$ and integrate $\frac{d v}{d x}$ to find $v$
e.g. $\begin{matrix} u = x & v = - \cos x \\ \frac{d u}{d x} = 1 & \frac{d v}{d x} = \sin x \end{matrix}$

STEP 3
Apply the integration by parts formula
e.g. $I = - x \cos x - \int_{}^{} - \cos x d x$

STEP 4
Work out the second integral, $\int_{}^{} v \frac{d u}{d x} d x$
Now include a “+c” (unless definite integration)
e.g. $I = - x \cos x + \sin x + c$

STEP 5
Simplify the answer if possible or apply the limits for definite integration
e.g. $I = \sin x - x \cos x + c$

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. $y = \ln x$ , $y = \arcsin x$ , $y = \arccos x$ , $y = \arctan x$
These can be integrated using parts however
- rewrite as the product ‘ $1 \times f (x)$ ’ and choose $u = f (x)$ and $\frac{d v}{d x} = 1$
- 1 is easy to integrate and the functions above have standard derivatives listed in the formula booklet

Exam Tip

If $\ln x$ or one of the inverse trig functions are one of the functions involved in the product then these should be assigned to " $u$ " when applying parts
- They are (realtively) easy to differentiate (to find $u'$ ) but are awkward to integrate

Worked example

a) Find $\int_{}^{} 5 x e^{3 x} d x$ .

5-9-2-ib-hl-aa-only-we2a-soltn

b) Show that $\int_{}^{} 8 x \ln x d x = 2 x^{2} (1 + \ln x^{2}) + c$ .

5-9-2-ib-hl-aa-only-we2b-soltn

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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 $x$

integration by parts will need to be applied again to the second integral

This occurs when one of the functions takes more than one derivative to become simple enough to make the second integral straightforward

These functions usually have the form $x^{2} g (x)$

How do I apply integration by parts more than once?

STEP 1
Name the integral if it doesn’t have one already!

STEP 2
Assign $u$ and $\frac{d v}{d x}$ . Find $\frac{d u}{d x}$ and $v$

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 $e^{x} \sin x$
- the derivative of $e^{x}$ is $e^{x}$
  - no matter how many times a function involving $e^{x}$ is differentiated, it will still involve $e^{x}$
- the derivative of $\sin x$ is $\cos x$
  - $\cos x$ would then have derivative $- \sin x$ , and so on
  - no matter how many times a function involving $\sin x$ or $\cos x$ is differentiated, it will still involve $\sin x$ or $\cos x$
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 ( $I$ ) at the start helps
- $I$ then appears twice in integration by parts
  - e.g. $I = g (x) - I$
    where $g (x)$ are parts of the integral not requiring further work
- It is then straightforward to rearrange and solve the problem
  - e.g. $2 I = g (x) + c$
    $I = \frac{1}{2} g (x) + c$

Worked example

a) Find

\int_{}^{} x^{2} \cos x d x

5-9-2-ib-hl-aa-only-we3a-soltn

b) Find

\int_{}^{} e^{x} \sin x d x

5-9-2-ib-hl-aa-only-we3b-soltn

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Further Techniques of Integration (DP IB Maths: AA HL)

Revision Note

What is integration by substitution?

How do I use integration by substitution?

What is integration by parts?

What is the formula for integration by parts?

How do I use integration by parts?

Can integration by parts be used when there is only a single function?

When will I have to repeat integration by parts?

How do I apply integration by parts more than once?

What if neither function ever becomes simpler when differentiating?

Author: Paul