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Substitution of u=tan(x/2)

  1. Jun 22, 2010 #1
    Hi,

    I've been doing some additional maths papers and I've seen the use of the substitution u=tan(x/2) in order to calculate integrals. In the mark scheme it states that this particular substitution used to be fairly common, however is not on the modern A-level syllabus.

    Would someone please mind advising me of suitable situations to use such a substitution? I am struggling to see when I should use it.


    Thanks,
    Oscar
     
  2. jcsd
  3. Jun 22, 2010 #2

    arildno

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    It is useful when your integrand consists of a ratio between to polynomials in of trignometric functions.

    For example:

    Let's look at:
    [tex]\int\frac{\cos\theta}{2\cos\theta-\sin\theta}d\theta[/tex]

    How would you integrate that one?

    Not very easy, but look at the following:
    [tex]\cos\theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}=\cos^{2}\frac{\theta}{2}(1-\tan^{2}\frac{\theta}{2})=\frac{1-u^{2}}{1+u^{2}}[/tex]
    [tex]\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=2\cos^{2}\frac{\theta}{2}\tan\frac{\theta}{2}=\frac{2u}{1+u^{2}}[/tex]

    Thus, we also have:
    [tex]\tan\theta=\frac{2u}{1-u^{2}}[/tex]

    In addition, we have:
    [tex]\frac{du}{d\theta}=\frac{1}{2}\frac{1}{\cos^{2}\frac{\theta}{2}}=\frac{1}{2}(u^{2}+1}\to{d\theta}=\frac{2du}{1+u^{2}}[/tex]

    Thus, the above integral can be converted into a rational expression of polynomials in the variable "u", and that can be solved using partial fractions decomposition. :smile:
     
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