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Integration of 1/(a^2sin^2(x)+b^2cos^2(x))^2)

  1. Dec 10, 2013 #1
    How do I integrate:
    [itex]\int\dfrac{dx}{(a^2sin^2(x)+b^2cos^2(x))^2}[/itex]
    Multiplying and dividing by [itex]sec^4(x)[/itex] doesn't work, neither does substituting [itex]tan[/itex].
    Any pointers would be appreciated.
     
    Last edited: Dec 10, 2013
  2. jcsd
  3. Dec 10, 2013 #2

    Simon Bridge

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    Use other trig identities to simplify the expression.

    $$a^2\sin^2x + b^2\cos^2x = \cos^2x[a^2\sec^2x +(b^2-a^2)]\\ \qquad =a^2+(b^2-a^2)\cos^2x\\ \qquad = (a^2-b^2)\sin^2x + b^2$$

    Also, consider similar triangles with the same angle x but different hypotenuses ... one has hypotenuse ##a^2## and the other ##b^2##

    Can you sub ##u^2=a^2\sin^2x+b^2\cos^2x##

    It's basically a matter of working through lots of things until you get a feel for the relationships.
     
    Last edited: Dec 10, 2013
  4. Dec 10, 2013 #3
    Thanks Simon, [itex](a^2−b^2)sin^2(x)+b^2[/itex] did help move forward but now I'm stuck at [itex]\int \dfrac{1+t^2}{(a^2t^2+b^2)^2} dt[/itex] And though I may be able to continue on from here trying this and that, isn't there a shorter way?
     
  5. Dec 10, 2013 #4
    The actual problem from where this stemmed out was this btw:
    [itex]\int_0^\pi\dfrac{x dx}{(a^2sin^2(x)+b^2cos^2(x))^2}[/itex]
    If it could've been solved without of all this, please tell me how.
     
  6. Dec 10, 2013 #5

    Simon Bridge

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    Have you tried partial fractions?
     
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