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Homework Help: Integration Question

  1. Oct 30, 2005 #1
    I'm stuck half-way through a question where I need to find the integral (with respect to x) of:

    tan(x/2).(cosec x + cot x)^2

    I've tried every method of integration that I know and I can't get it. Is it actually possible to integrate the expression?

    If it's any help, this problem is part of the further maths A-level and the original problem was:

    Find the general solution of the differential equation dy/dx - 2y.cosec x = tan(x/2), 0<x<pi

    I got to the integral I'm stuck on by using an integrating factor of (cosec x + cot x)^2, so that the LHS became the derivative of y(cosec x + cot x)^2 (and therefore y(cosec x + cot x)^2 = integral[tan(x/2).(cosec x + cot x)^2]dx ).

    P.S. Sorry if the above isn't very clear, since I don't know how to get mathematical symbols on these boards.
     
  2. jcsd
  3. Oct 30, 2005 #2

    benorin

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    The value of that integral is 2ln(sin(x/2))+C

    All you need to know is the right trig identity: [tex]\tan \left(\frac{x}{2}\right)=\frac{\sin(x)}{\cos(x)+1}[/tex].
     
  4. Oct 30, 2005 #3

    Fermat

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    The integating factor simplifies by using half-angle formulae.

    sinx = 2sin(x/2).cos(x/2)
    cos(x) = 2cosĀ²(x/2) - 1

    btw, you can get maths symbols by using Latex
     
  5. Oct 30, 2005 #4

    benorin

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    Here's the integral

    Use [tex]\tan\left(\frac{x}{2}\right)=\frac{\sin(x)}{\cos(x)+1} [/tex] like this:
    [tex]\int\tan\left(\frac{x}{2}\right)\left(\csc(x) + \cot(x)\right)^{2}dx = \int\left(\frac{\sin(x)}{\cos(x)+1}\right)\left(\frac{1}{\sin(x)} + \frac{\cos(x)}{\sin(x)}\right)^{2}dx[/tex]
    [tex]= \int\left(\frac{\sin(x)}{\cos(x)+1}\right)\left(\frac{\cos(x)+1}{\sin(x)}\right)^{2}dx = \int\frac{\cos(x)+1}{\sin(x)}dx = \int\cot\left(\frac{x}{2}\right)dx[/tex]
    Let [tex]u=\frac{x}{2}\Rightarrow 2du=dx[/tex] to get
    [tex] 2\int\cot(u)du = 2\int\frac{\cos(u)}{\sin(u)}du[/tex]
    set [tex]w=\sin(u)\Rightarrow dw=\cos(u)du[/tex] which yeilds
    [tex]2\int\frac{dw}{w} = 2\ln(w) + C = 2\ln(\sin(u)) + C = 2\ln\left(\sin\left(\frac{x}{2}\right)\right) + C[/tex]
    There, all done.:rolleyes:
     
    Last edited: Oct 30, 2005
  6. Nov 3, 2005 #5
    Thanks guys. You've been very helpful.
     
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