Integrating Further Functions (DP IB Maths: AA HL)

Revision Note

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Author

Paul

Last updated

1 December 2023

As with other problems in integration the results in this revision note may have further uses such as

evaluating a definite integral
finding the constant of integration
finding areas under a curve, between a line and a curve or between two curves

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Integrating with Reciprocal Trigonometric Functions

cosec (cosecant, csc), sec (secant) and cot (cotangent) are the reciprocal functions of sine, cosine and tangent respectively.

What are the antiderivatives involving reciprocal trigonometric functions?

$\int_{}^{} \sec^{2} x d x = \tan x + c$
$\int_{}^{} \sec x \tan x d x = \sec x + c$
$\int_{}^{} cosec x \cot x d x = - cosec x + c$
$\int_{}^{} {cosec}^{2} x d x = - \cot x + c$
These are not given in the formula booklet directly

they are listed the other way round as ‘standard derivatives’
be careful with the negatives in the last two results
and remember “+c” !

**How do I integrate these if a linear function of x is involved?**

All integration rules could apply alongside the results above
The use of reverse chain rule is particularly common
- For linear functions the following results can be useful
  - $\int_{}^{} \sec^{2} (a x + b) d x = \frac{1}{a} \tan (a x + b) + c$
  - $\int_{}^{} \sec (a x + b) \tan (a x + b) d x = \frac{1}{a} \sec (a x + b) + c$
  - $\int_{}^{} cosec (a x + b) \cot (a x + b) d x = - \frac{1}{a} cosec (a x + b) + c$
  - $\int_{}^{} {cosec}^{2} (a x + b) d x = - \frac{1}{a} \cot (a x + b) + c$
- These are not in the formula booklet
  - they can be deduced by spotting reverse chain rule
  - they are not essential to remember but can make problems easier

Examiner Tip

Even if you think you have remembered these antiderivatives, always use the formula booklet to double check
- those squares, negatives and "1 over"'s are easy to get muddled up!
Remember to use 'adjust' and 'compensate' for reverse chain rule when coefficients are involved

Worked example

The graph of $y = f (x)$ where $f (x) = \int_{}^{} 2 \sec^{2} 5 x d x$ passes through the point $(\frac{π}{3}, 0)$ .

Show that $5 y = 2 (\sqrt{3} + \tan 5 x)$ .

5-9-1-ib-hl-aa-only-we1-soltn

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Integrating with Inverse Trigonometric Functions

arcsin, arccos and arctan are (one-to-one) functions defined as the inverse functions of sine, cosine and tangent respectively.

What are the antiderivatives involving the inverse trigonometric functions?

$\int_{}^{} \frac{1}{\sqrt{1 - x^{2}}} d x = \arcsin x + c$
$\int_{}^{} \frac{1}{1 + x^{2}} d x = \arctan x + c$
Note that the antiderivative involving $\arccos x$ would arise from

$\int_{}^{} - \frac{1}{\sqrt{1 - x^{2}}} d x = \arccos x + c$

- However, the negative can be treated as a coefficient of -1 and so

$\int_{}^{} - \frac{1}{\sqrt{1 - x^{2}}} d x = - \int_{}^{} \frac{1}{\sqrt{1 - x^{2}}} d x = - \arcsin x + c$

- Similarly,

$\int_{}^{} \frac{1}{\sqrt{1 - x^{2}}} d x = - \int_{}^{} - \frac{1}{\sqrt{1 - x^{2}}} d x = - \arccos x + c$

Unless a question requires otherwise, stick to the first two results
These are listed in the formula booklet the other way round as ‘standard derivatives’
For the antiderivative involving $\arctan x$ , note that $(1 + x^{2})$ is the same as $(x^{2} + 1)$

How do I integrate these expressions if the denominator is not in the correct form?

Some problems involve integrands that look very similar to the above
- but the denominators start with a number other than one
- there are three particular cases to consider
The first two cases involve denominators of the form $a^{2} \pm {(b x)}^{2}$ (with or without the square root!)
- In the case $b = 1$ (i.e. denominator of the form $a^{2} \pm x^{2}$ ) there are two standard results
  - $\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \arctan (\frac{x}{a}) + c$
  - $\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \arcsin (\frac{x}{a}) + c, | x | < a$
  - Both of these are given in the formula booklet
  - Note in the first result, $a^{2} + x^{2}$ could be written $x^{2} + a^{2}$
- In cases where $b \neq 1$ then the integrand can be rewritten by taking a factor of $a^{2}$
  - the factor will be a constant that can be taken outside the integral
  - the remaining denominator will then start with 1
  - e.g. $9 + 4 x^{2} = 9 (1 + \frac{4}{9} x^{2}) = 9 (1 + {(\frac{2}{3} x)}^{2})$
The third type of problem occurs when the denominator has a (three term) quadratic
- i.e. denominators of the form $a x^{2} + b x + c$
  (a rearrangement of this is more likely)
  - the integrand can be rewritten by completing the square
  - e.g. $5 - x^{2} + 4 x = 5 - (x^{2} + 4 x) = 5 - [{(x + 2)}^{2} - 4] = 9 - {(x + 2)}^{2}$
    This can then be dealt with like the second type of problem above with " $x$ " replaced by " $x + 2$ "
  - This works since the derivative of $x + 2$ is the same as the derivative of $x$
    There is essentially no reverse chain rule to consider

Examiner Tip

Always start integrals involving the inverse trig functions by rewriting the denominator into a recognisable form
- The numerator and/or any constant factors can be dealt with afterwards, using 'adjust' and 'compensate' if necessary

Worked example

a) Find

\int_{}^{} \frac{1}{9 + x^{2}} d x

5-9-1-ib-hl-aa-only-we2a-soltn

b) Find $\int_{}^{} \frac{1}{\sqrt{5 - x^{2} + 4 x}} d x$ .

5-9-1-ib-hl-aa-only-we2b-soltn

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Integrating Exponential & Logarithmic Functions

Exponential functions have the general form $y = a^{x}$ . Special case: $y = e^{x}$ .

Logarithmic functions have the general form $y = \log_{a} x$ . Special case: $y = \log_{e} x = \ln x$ .

What are the antiderivatives of exponential and logarithmic functions?

Those involving the special cases have been met before
- $\int_{}^{} e^{x} d x = e^{x} + c$
- $\int_{}^{} \frac{1}{x} d x = \ln | x | + c$
- These are given in the formula booklet
Also
- $\int_{}^{} a^{x} d x = \frac{1}{\ln a} a^{x} + c$
- This is also given in the formula booklet
By reverse chain rule
- $\int_{}^{} \frac{1}{x \ln a} d x = \log_{a} | x | + c$
- This is not in the formula booklet
  - but the derivative of $\log_{a} x$ is given
There is also the reverse chain rule to look out for
- this occurs when the numerator is (almost) the derivative of the denominator
- $\int_{}^{} \frac{f' (x)}{f (x)} d x = \ln |f (x)| + c$

**How do I integrate exponentials and logarithms with a linear function of x involved?**

For the special cases involving $e$ and $\ln$
- $\int_{}^{} e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$
- $\int_{}^{} \frac{1}{a x + b} d x = \frac{1}{a} \ln |a x + b| + c$
For the general cases
- $\int a^{p x + q} d x = \frac{1}{p \ln a} a^{p x + q} + c$
- $\int \frac{1}{(p x + q) \ln a} d x = \frac{1}{p} \log_{a} | p x + q | + c$
These four results are not in the formula booklet but all can be derived using ‘adjust and compensate’ from reverse chain rule

Examiner Tip

Remember to always use the modulus signs for logarithmic terms in the antiderivative
- Once it is deduced that $g (x)$ in $\ln | g (x) |$ , say, is guaranteed to be positive, the modulus signs can be replaced with brackets