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Homework Help: Couple of Problems - simple integrals

  1. Oct 9, 2006 #1
    Please confirm my answers

    1) definite integral 0 to 1 [1/1+x^2 + 9e^x - 1/sqrt(x)]
    = tan^-1(x) + 9e^x - 2x^1/2
    I know that to evaluate it I just plug in b for x - a for x

    2) indefinte integral sin(3t)cos(3t)

    = 1/3 (-cos(3t)sin(3t))dt
     
  2. jcsd
  3. Oct 9, 2006 #2

    dextercioby

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    2) is definitely wrong. As for 1), i think you mean the simpler version, but from what you've written one understands

    [tex] \int_{0}^{1}\frac{dx}{1+x^{2}+9e^{x}-\frac{1}{\sqrt{x}}} [/tex]

    Daniel.
     
  4. Oct 9, 2006 #3
    hi, for #1 I meant the integral from 0 to 1 [(1/1+x^2)+ 9e^x + (1/sqrt(x))]dx

    sorry, I'm not sure how to do edit it as well as you did.

    Where exactly did I mess up on #2 and how should I approach it?

    Thanks!
     
  5. Oct 9, 2006 #4

    dextercioby

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    HINT: [itex](\sin 3x)'=3\cos 3x [/itex]. So i guess a substitution will do.

    Daniel.
     
  6. Oct 9, 2006 #5
    thanks alot for the hint it was very helpful

    now I get: 1/6(sin3x)^2+C

    please let me know if this is correct.

    Thanks again.
     
  7. Oct 9, 2006 #6

    dextercioby

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    Surely it is. You can always check the result by differentiation.

    Daniel.
     
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