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Integerating (1/((sinx)^2))

  1. Mar 21, 2009 #1
    1. The problem statement, all variables and given/known data

    25[tex]\int\frac{1}{sin^{2}x}[/tex]dx


    3. The attempt at a solution

    I wasn't sure if I could change [tex]\frac{1}{sin^{2}x}[/tex] to Csc[tex]^{2}[/tex]x but when I did I ended up with -25Cotx which when I checked the integral in my calculator and it was wrong. So now i'm lost...
     
  2. jcsd
  3. Mar 21, 2009 #2
    FYI, the original problem was [tex]\int\frac{1}{x^{2}\sqrt{25-x^{2}}}[/tex]dx I used trigonometric substitution to get to the problem above.
     
  4. Mar 21, 2009 #3
    Well in an online integral table I found that [tex]\int \csc^2 ax dx = -\frac{1}{a} \cot ax + C[/tex] so you should probably come up with [tex]-\frac{cot(25x)}{25} + C[/tex].
     
  5. Mar 21, 2009 #4
    -25Cot(x) is the integral of 25/sin^2(x), you can check it by computing the derivative.
     
  6. Mar 21, 2009 #5
    Well did I go wrong before I got to the sin integral? because when I checked it against the original problem it didn't match.
     
  7. Mar 21, 2009 #6
    Oh I did catch that it should be 1/25 not 25. But its still slightly off.
     
  8. Mar 21, 2009 #7

    Dick

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    Judging from what you've shown us, you used the substitution x=5*sin(u) to reduce the integral to (1/25) times the integral (1/sIn(u)^2)*du. That's fine. So you've got -cot(u)/25 as the integral. You still have to express that in terms of x.
     
  9. Mar 21, 2009 #8
    I substituted [tex]\frac{\sqrt{25-x^{2}}}{x}[/tex] for cot So what I end up with is -[tex]\frac{\sqrt{25-x^{2}}}{25x}[/tex]+c
     
  10. Mar 22, 2009 #9

    Dick

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    That looks fine to me.
     
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