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Integration Problem

  1. Jan 11, 2006 #1
    Hey everyone,
    I can't seem to figure out how to do this integral. If anyone could help that would be great. I would appreciate some hints/advice, rather than just the answer. Thanks!
    The question is:
    Evaluate the indefinite integral:
    Ok... So i know you can take the integral of each term seperatly because it is a sum and that is a principle of integrals.... so i get the first part to be -4cos(x). However I am not sure how to evalue the antiderivative of 3tan(x). If anyone could help that would be great!
    Thanks again.
  2. jcsd
  3. Jan 11, 2006 #2


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    Noting that tan x = sin x/cos x, you can integrate by subsitution using u=cos x.
  4. Jan 13, 2006 #3
    ok...... so i tried that.
    so with the substitution you end up with du/u whose antiderivative is -ln(absolute value)U. U is cosx so it is -ln(abs)cosx. Moving the negative sign into the logarithm gives ln(abs)secx. So my final answer would be...
    4cosx+3ln(abs)secx. Im not to sure this is right though. Could one of you smart math people check and if i made an error somewhere point me in the right direction.
    Thanks alot for the help!
  5. Jan 13, 2006 #4


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    Watch your signs, the antiderivative of sin(x) is -cos(x).
  6. Jan 13, 2006 #5
    Oh yes...
    So is the final answer:
    Thanks for pointing that out. (i think i fixed it)

    Another quick question that I cant seem to figure out...
    Find the definite integral from 0 to 1 of:
    I thought this may be another substitution question but I cant see what to substitute. Any help would be great!
    Last edited: Jan 13, 2006
  7. Jan 13, 2006 #6


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    I was referring to the term 4cos(x) which should actually be -4cos(x) :smile:
  8. Jan 13, 2006 #7


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    For the second, let u= 5x + 6 from there it is easy to see that x^2 = (u-6)^2/25, then expand and integrate.
  9. Jan 14, 2006 #8
    I dont really understand that Tx. If i let U=5xt6 i end up with x^2(U)^(1/2)(du/5) which doesnt seem to help me in any way. Some more advice would be good! :smile: I appreciate the help!
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