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A trigonometric integration

  1. May 20, 2012 #1
    Hi everyone,

    In my research project, i am struck with an integration. Can someone help me out:


  2. jcsd
  3. May 20, 2012 #2
    Hey Sandeep.

    How did you attempt this integral? It will be easier(and ethical) to help you out if you share how you worked on it. :smile:
  4. May 20, 2012 #3
    I tried with substitution √k*sinx = t;

    But it ends up with ∫dt/[(1 - t2/k)*(1 - t2)]

    I dont know how to go ahead with this.

    I tried with ksin2x = t also. But that too doesnt work.

    Please help me. This is one of the last parts of a huge integration i am doing...

  5. May 20, 2012 #4
    Try breaking [itex]1-ksin^{2}x[/itex] into [itex](1-\sqrt{k}sin x)(1+\sqrt{k}sinx)[/itex]

    Now try splitting the term into two simple integrals of the form [itex]\frac{1}{a+bsinx}[/itex]
  6. May 20, 2012 #5
    But it is (1 - ksin2x).

    So, how can we split it into two simple integrals?

  7. May 20, 2012 #6
  8. May 20, 2012 #7
    Thank you.

    Will Go through it..

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