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Stuck with an integral

  1. Jun 18, 2007 #1
    Hello everyone

    I need some help in performing the following integration (not HW):

    [tex]\int_{0}^{\pi/4}\left(\frac{x}{x\sin x + \cos x}\right)^{2}dx[/tex]

    I tried integration by parts, but it leads nowhere. Any suggestions would be appreciated.

    PS--Mathematica gives the answer as [itex](4-\pi)/(4+\pi)[/itex] but is unable to perform the integration with the [itex]x^2[/itex] term in the numerator replaced by unity.
  2. jcsd
  3. Jun 18, 2007 #2
    To solve your integral, you can start by differentiating

    [tex]-\frac{x \sec{x}}{(x \sin{x} + \cos{x})}[/tex]

    this will give you

    [tex]-(\frac{x \sec{x} }{x \sin{x} + \cos{x}})^{\prime} = \frac{x^2}{(x \sin{x} + \cos{x})^2} - \frac{(\sec{x} + x \sec{x} \tan{x})}{(x \sin{x} + \cos{x})}[/tex]

    Now, you'll recognize the first term on the right hand side as your integrand. To evaluate the second term on the right hand side, you can first take the [tex]\sec{x}[/tex] out of the top half to get

    [tex]\frac{(\sec{x} + x \sec{x} \tan{x})}{(x \sin{x} + \cos{x})} = \sec{x} \frac{1 + x \tan{x}}{(x \sin{x} + \cos{x})}[/tex]

    Now, divide the bottom half by [tex]\cos{x}[/tex] to get

    [tex]\frac{\sec{x}}{\cos{x}} \frac{(1 + x \tan{x})}{(x \tan{x} + 1)} = \sec^2{x}[/tex]

    Now, the integral of [tex]\sec^2{x}[/tex] is

    [tex]\int{\sec^2{x} dx} = \tan{x} + C [/tex]

    therefore, your integral, is

    [tex]\int{(\frac{x}{x \sin x + \cos x})^{2} dx} = \tan{x} - \frac{x \sec{x}}{(x \sin{x} + \cos{x})} + C[/tex]

    where C is a constant of integration.
    Last edited: Jun 18, 2007
  4. Jun 19, 2007 #3

    Gib Z

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    Homework Helper

    I'm not sure about you Matthew but most people don't have a preset function they know the should differentiate and compare their integral to. Otherwise one might just say to differentiate [tex]\frac{\sin x - x\cos x}{\cos x - x\sin x}[/tex] and see what you get.
  5. Jun 19, 2007 #4
    Fair enough, but it wasn't a guess -- I got there by trying to find out what function, f(x), when divided by [tex](x \sin{x} + \cos{x})[/tex] yields the integrand in part of its derivative. (The answer of course is [tex]f(x) = -x \sec{x}[/tex]). Then it's a question of seeing if you can integrate the other part(s) of the derivative -- if you can, you have a solution. In this case, it was possible.
    Last edited: Jun 19, 2007
  6. Jun 19, 2007 #5

    Gib Z

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    Homework Helper

    In that case, good work mate :)
  7. Jun 19, 2007 #6
    No, you're right -- I should have made this clear at the start. Sorry folks.
  8. Jun 20, 2007 #7
    Thanks Matthew and GibZ

    Sorry for the late acknowledgement...I figured out how to do it, by a method similar to that suggested.

    GibZ, if I take your function with the minus replaced by plus in the denominator, then the derivative equals the integrand. So thats an equivalent way of doing it.

    Thanks again
    Last edited: Jun 20, 2007
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