Integrating with Inverse Trigonometric Functions (DP IB Analysis & Approaches (AA)): Revision Note

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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 Tips and Tricks

  • 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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