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Evaluate this integral

  1. Jun 23, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex] \displaystyle \int^{\pi /2}_0 \dfrac{xcosxsinx}{(a^2 cos^2 x+b^2 sin^2 x)^2} dx [/itex]

    2. Relevant equations

    3. The attempt at a solution

    2I = [itex] \displaystyle \int^{\pi /2}_0 \dfrac{xcosxsinx}{(a^2 cos^2 x+b^2 sin^2 x)^2} + \dfrac{(\pi /2 -x)cosxsinx}{(a^2 sin^2 x+b^2 cos^2 x)^2} dx [/itex]

    How to simplify after this?
     
  2. jcsd
  3. Jun 23, 2013 #2
    I couldn't think of a smarter way but the following should work.

    You can write the given integral as
    [tex]\int_0^{\pi/2} \frac{x\tan x \sec^2x}{(a^2+b^2\tan^2x)^2}dx[/tex]
    Use the substitution ##\tan x=t## and then integrate by parts.

    And what about your previous thread? Did you get the answer? :rolleyes:
     
  4. Jun 23, 2013 #3

    lurflurf

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    ^Are you trying to get the antiderivative? It is easy to get rid of the x in front by integrating by parts, but the new integral is hard. Perhaps the bounds can be used, but the a and b break the symmetry.
     
  5. Jun 23, 2013 #4
    Yes, I am trying to find the antiderivative. The new integral isn't hard but only a bit lengthy (I think).

    The new integral after substitution,
    [tex]\int_0^{\infty} \frac{\arctan t\cdot t}{(a^2+b^2t^2)^2}dt[/tex]

    This can be solved using integration by parts.
     
  6. Jun 24, 2013 #5

    utkarshakash

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    Yes I got the answer of previous question.
     
  7. Jun 24, 2013 #6
    I found a shorter method.

    Solve the given integral using integration by parts with ##u=x## and ##v## be the rest i.e
    [tex]x\int \frac{\sin x \cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx-\int \left(\int \frac{\sin x \cos x}{(a^2\cos^2x+b^2\sin^2x)^2}dx\right)dx[/tex]

    Use the substitution, ##(a^2\cos^2x+b^2\sin^2x)=t## that should give the answer.
     
  8. Jun 24, 2013 #7
    This question is very easy by utkarshakash's method but he got afraid to see the big size of his integrand. :p

    I too at first thought would use his method.

    After simplifying you get,

    2I=....


    This will eliminate x. No need to integrate by parts. :)

    Hint: Then use the substitution as Pranav suggested. Take sqrt(denominator)=t...
     
    Last edited: Jun 24, 2013
  9. Jun 24, 2013 #8
    No. How do you conclude that?
     
  10. Jun 25, 2013 #9
    Isn't is obvious ?

    Let

    I=π/20∫{xcosxsinx}/{(a2 cos2 x+b2 sin2 x)2} dx

    Then applying the property of definite integration,

    I=π/20∫{(π/2-x)cosxsinx}/{(a2 cos2 x+b2 sin2 x)2} dx

    On adding,

    2I= I=π/20∫{π/2cosxsinx}/{(a2 cos2 x+b2 sin2 x)2} dx

    Now take, a2 cos2 x+b2 sin2 x=t
     
  11. Jun 25, 2013 #10
    [tex]\cos(\pi/2-x)=\sin x[/tex]
     
    Last edited: Jun 25, 2013
  12. Jun 25, 2013 #11
    So what ?

    cos x changes to sin x but sin x also changes to cos x. The net result remains, what I typed in my previous post. I have done such type of questions.
     
  13. Jun 25, 2013 #12
    You ignored the denominator when you applied the property.
     
  14. Jun 25, 2013 #13

    utkarshakash

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    No. You are going the wrong way. Note that though cos x and sin x interchange, the coefficents are also reversed. [itex] a^2 cos^2 x + b^2 sin^2 x[/itex] changes to [itex] a^2 sin^2 x + b^2 cos^2 x[/itex]. Now you can't just add the numerator simply. You know what I mean.
     
  15. Jun 25, 2013 #14
    Did you arrive at the answer? :smile:
     
  16. Jun 25, 2013 #15

    utkarshakash

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    I got the answer by your method. Thanks!
     
  17. Jun 25, 2013 #16
    Glad to help!
     
  18. Jun 26, 2013 #17
    Oh, I ignored the denominator when I applied the property !! :redface:

    However, when you integrate by parts, you can also divide both sides by cos4x, and then take tan2(x) = t. This will simplify the integrand to a mere identity.
     
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