Integrating Further Functions (DP IB Maths: AA HL)

Revision Note

Paul

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Paul

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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?

  • integral subscript blank superscript blank sec squared space x space d x equals tan space x plus c
  • integral subscript blank superscript blank sec space x space tan space x space d x equals sec space x plus c
  • integral subscript blank superscript blank cosec space x space cot space x space d x equals negative cosec space x plus c
  • integral subscript blank superscript blank cosec squared space x space d x equals negative cot space x plus 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
      • integral subscript blank superscript blank sec squared open parentheses a x plus b close parentheses space straight d x equals 1 over a tan open parentheses a x italic plus b close parentheses plus straight c
      • integral subscript blank superscript blank sec open parentheses a x plus b close parentheses space tan open parentheses a x plus b close parentheses space d x equals 1 over a sec open parentheses a x plus b close parentheses plus straight c
      • integral subscript blank superscript blank cosec open parentheses a x plus b close parentheses space cot open parentheses a x plus b close parentheses space d x equals negative 1 over a cosec open parentheses a x plus b close parentheses plus straight c
      • integral subscript blank superscript blank cosec squared open parentheses a x plus b close parentheses space d x equals negative 1 over a cot open parentheses a x plus b close parentheses plus straight 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 equals f left parenthesis x right parenthesis where f left parenthesis x right parenthesis equals integral subscript blank superscript blank 2 sec squared space 5 x space d x passes through the point open parentheses pi over 3 comma space 0 close parentheses.

Show that 5 y equals 2 open parentheses square root of 3 plus tan space 5 x close parentheses.

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?

  • integral subscript blank superscript blank fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x equals arcsin space x plus c
  • integral subscript blank superscript blank fraction numerator 1 over denominator 1 plus x squared end fraction space d x equals arctan space x plus c
  • Note that the antiderivative involving arccos space x would arise from

integral subscript blank superscript blank minus fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x equals arccos space x plus c

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

integral subscript blank superscript blank minus fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x equals negative integral subscript blank superscript blank fraction numerator 1 over denominator square root of 1 minus x squared end root space end fraction space d x equals negative arcsin space x plus c

    • Similarly,

integral subscript blank superscript blank fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x equals negative integral subscript blank superscript blank minus fraction numerator 1 over denominator square root of 1 minus x squared end root end fraction space d x equals negative arccos space x plus 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 space x, note that open parentheses 1 plus x squared close parentheses is the same as open parentheses x squared plus 1 close parentheses

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 squared plus-or-minus open parentheses b x close parentheses squared space (with or without the square root!)
    • In the case bold italic b bold equals bold 1 (i.e. denominator of the form a squared plus-or-minus x squared) there are two standard results
      • integral fraction numerator 1 over denominator a squared plus x squared end fraction space straight d x equals 1 over a arctan open parentheses x over a close parentheses space plus c
      • integral fraction numerator 1 over denominator square root of a squared minus x squared end root end fraction space straight d x equals arcsin open parentheses x over a close parentheses plus c comma space space vertical line x vertical line less than a
      • Both of these are given in the formula booklet
      • Note in the first result, a squared plus x squared could be written x squared plus a squared
    • In cases where bold italic b bold not equal to bold 1 then the integrand can be rewritten by taking a factor of a squared
      • the factor will be a constant that can be taken outside the integral
      • the remaining denominator will then start with 1
      • e.g. 9 plus 4 x squared equals 9 open parentheses 1 plus 4 over 9 x squared close parentheses equals 9 open parentheses 1 plus open parentheses 2 over 3 x close parentheses squared close parentheses
  • The third type of problem occurs when the denominator has a (three term) quadratic
    • i.e.  denominators of the form bold italic a bold italic x to the power of bold 2 bold plus bold italic b bold italic x bold plus bold italic c
      (a rearrangement of this is more likely)
      • the integrand can be rewritten by completing the square
      • e.g. 5 minus x squared plus 4 x equals 5 minus open parentheses x squared plus 4 x close parentheses equals 5 minus open square brackets open parentheses x plus 2 close parentheses squared minus 4 close square brackets equals 9 minus open parentheses x plus 2 close parentheses squared
        This can then be dealt with like the second type of problem above with "x" replaced by "x plus 2"
      • This works since the derivative of x plus 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 integral subscript blank superscript blank fraction numerator 1 over denominator 9 plus x squared end fraction space d x.

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

b)       Find integral subscript blank superscript blank fraction numerator 1 over denominator square root of 5 minus x squared plus 4 x end root end fraction space 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 equals a to the power of x.  Special case:  y equals straight e to the power of x.

Logarithmic functions have the general form y equals log subscript a x.  Special case:  y equals log subscript straight e x equals ln space x.

What are the antiderivatives of exponential and logarithmic functions?

  • Those involving the special cases have been met before
    • integral subscript blank superscript blank straight e to the power of x space d x equals straight e to the power of x plus c
    • integral subscript blank superscript blank 1 over x space d x equals ln space vertical line x vertical line plus c
    • These are given in the formula booklet
  • Also
    • integral subscript blank superscript blank a to the power of x space d x equals fraction numerator 1 over denominator ln space a end fraction a to the power of x plus c
    • This is also given in the formula booklet
  • By reverse chain rule
    • integral subscript blank superscript blank fraction numerator 1 over denominator x ln space a end fraction space d x equals log subscript a vertical line x vertical line plus c
    • This is not in the formula booklet
      • but the derivative of log subscript 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
    • integral subscript blank superscript blank fraction numerator f apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction space d x equals ln space open vertical bar space f left parenthesis x right parenthesis space close vertical bar plus c

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

  • For the special cases involving e and ln
    • integral subscript blank superscript blank straight e to the power of a x plus b end exponent space d x equals 1 over a straight e to the power of a x plus b end exponent plus straight c
    • integral subscript blank superscript blank fraction numerator 1 over denominator a x plus b end fraction space d x equals 1 over a ln space open vertical bar space a x plus b close vertical bar plus c
  • For the general cases
    • integral a to the power of space p x plus q end exponent space straight d x equals fraction numerator 1 over denominator p space ln space a end fraction a to the power of space p x plus q end exponent plus c
    • integral fraction numerator 1 over denominator open parentheses p x plus q close parentheses ln space a end fraction space straight d x equals 1 over p log subscript a vertical line p x plus q vertical line plus 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 left parenthesis x right parenthesis in  ln space vertical line g left parenthesis x right parenthesis vertical line, say, is guaranteed to be positive, the modulus signs can be replaced with brackets

Worked example

a)       Show that integral subscript 1 superscript 2 4 to the power of x space d x equals fraction numerator 6 over denominator ln space 2 end fraction.

5-9-1-ib-hl-aa-only-we3a-soltn

b)       Find integral fraction numerator 1 over denominator open parentheses 2 x minus 1 close parentheses space ln space 3 end fraction space straight d x.

zgf4~aK9_5-9-1-ib-hl-aa-only-we3b-soltn

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Paul

Author: Paul

Expertise: Maths

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.