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Indefinite integral of csc()

  1. Jul 8, 2014 #1
    What is the indefinite integral of [itex]cosec(\theta)[/itex]?
     
  2. jcsd
  3. Jul 8, 2014 #2

    maajdl

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  4. Jul 8, 2014 #3

    HallsofIvy

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    [itex]cosc(x)= \frac{1}{sin(x)}[/itex].

    [tex]\int cosec(x)dx= \int \frac{1}{sin(x)}dx= \int\frac{sin(x)}{sin^2(x)}dx= \int\frac{sin(x)}{1- cos^2(x)}dx[/tex].

    Now let u= cos(x) so that [itex]du= -sin(x)dx[/itex].
     
  5. Jul 8, 2014 #4

    verty

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    If you know the trick to integrate ##sec(x)##, try it with the co-functions instead.
     
  6. Jul 8, 2014 #5
    Here's an alternative solution: $$\eqalign{
    \int\csc x\,\mathrm dx &= \int\left(\csc x\dfrac{\csc x-\cot x}{\csc x-\cot x}\right)\mathrm dx \\
    &=\int\left(\dfrac{\csc^2 x-\cot x\csc x}{\csc x-\cot x}\right)\mathrm dx.
    }$$
    Now use the [itex]u[/itex]-substitution [itex]u=\csc x-\cot x[/itex] and you'll get: $$\int\dfrac{1}{u}\mathrm du=\ln|u|+{\cal C}=\ln|\csc x-\cot x|+{\cal C}.$$
     
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