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IBMathsDPApplications & Interpretation (AI)HLRevision Notes5. CalculusFurther IntegrationIntegrating Special Functions

Integrating Special Functions (DP IB Applications & Interpretation (AI)): Revision Note

Roger B

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Roger B

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Integrating Trig Functions

How do I integrate sin, cos and 1/cos2?

  • The antiderivatives for sine and cosine are

bold space bold integral bold sin bold space bold italic x bold space bold d bold italic x bold equals bold minus bold cos bold space bold italic x bold plus bold italic c

bold space bold integral bold cos bold space bold italic x bold space bold d bold italic x bold equals bold sin bold space bold italic x bold plus bold italic c

wherebold space bold italic c is the constant of integration

  • Also, from the derivative ofspace tan space x

bold space bold integral fraction numerator bold 1 over denominator bold cos to the power of bold 2 bold space bold italic x end fraction bold space bold d bold italic x bold equals bold tan bold space bold italic x bold plus bold italic c

  • All three of these standard integrals are in the formula booklet

  • For the linear functionbold space bold italic a bold italic x bold plus bold italic b, wherespace bold italic a andspace bold italic b are constants,

bold space bold integral bold sin bold space bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis bold space bold d bold italic x bold equals bold minus bold 1 over bold italic a bold cos bold space bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis bold plus bold italic c

bold space bold integral bold cos bold space bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis bold space bold d bold italic x bold equals bold 1 over bold italic a bold sin bold space bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis bold plus bold italic c

bold space bold integral fraction numerator bold 1 over denominator bold cos to the power of bold 2 bold space bold left parenthesis bold italic a bold italic x bold italic plus bold italic b bold right parenthesis end fraction bold space bold d bold italic x bold equals bold 1 over bold italic a bold tan bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis bold plus bold italic c

  • For calculus with trigonometric functions angles must be measured in radians

    • Ensure you know how to change the angle mode on your GDC

Examiner Tips and Tricks

  • Make sure you have a copy of the formula booklet during revision but don't try to remember everything in the formula booklet

    • However, do be familiar with the layout of the formula booklet

      • You’ll be able to quickly locate whatever you are after

      • You do not want to be searching every line of every page!

    • For formulae you think you have remembered, use the booklet to double-check

Worked Example

a) Find, in the formspace straight F left parenthesis x right parenthesis plus c, an expression for each integral

  1. space integral cos space x space straight d x

  2. space integral fraction numerator 1 over denominator cos squared space open parentheses 3 x minus begin display style straight pi over 3 end style close parentheses end fraction space straight d x

5-4-1-ib-hl-ai-aa-extraaa-ai-we1a-soltn

b)       A curve has equationspace y equals integral 2 sin open parentheses 2 x plus straight pi over 6 close parentheses space straight d x.

The curve passes through the point with coordinatesspace open parentheses straight pi over 3 comma space square root of 3 close parentheses.

Find an expression forspace y.

5-4-1-ib-hl-ai-aa-extraaa-we1b-soltn-

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Integrating e^x & 1/x

How do I integrate exponentials and 1/x?

  • The antiderivatives involvingbold space bold e to the power of bold italic x andspace bold ln bold space bold italic x are

bold space bold integral bold space bold e to the power of bold italic x bold space bold d bold italic x bold equals bold space bold e to the power of bold italic x bold plus bold italic c

where bold italic c is the constant of integration

  • These are given in the formula booklet

  • For the linear functionbold space bold left parenthesis bold italic a bold italic x bold plus bold italic b bold right parenthesis, wherespace bold italic a andspace bold italic b are constants,

 bold space bold integral bold e to the power of bold italic a bold italic x bold italic plus bold italic b end exponent bold space bold d bold italic x bold equals bold 1 over bold italic a bold e to the power of bold italic a bold italic x bold italic plus bold italic b end exponent bold plus bold italic c

  • It follows from the last result that

 

  • which can be deduced using Reverse Chain Rule

  • With ln, it can be useful to write the constant of integration,space c, as a logarithm

    • using the laws of logarithms, the answer can be written as a single term

    • wherespace k is a constant

    • This is similar to the special case of differentiatingspace ln space left parenthesis a x plus b right parenthesis whenspace b equals 0

Examiner Tips and Tricks

  • When revising, familiarise yourself with the layout of this section of the formula booklet, make sure you know what is and isn't in there and how to find it very quickly

Worked Example

A curve has the gradient functionspace f apostrophe left parenthesis x right parenthesis equals fraction numerator 3 over denominator 3 x plus 2 end fraction plus straight e to the power of 4 minus x end exponent.


Given the exact value ofspace f left parenthesis 1 right parenthesis isspace ln space 10 minus straight e cubed find an expression forspace f left parenthesis x right parenthesis.

NA5HYQ75_5-4-1-ib-sl-aa-only-we2-soltn

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    Roger B

    Author: Roger B

    Expertise: Maths

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.