Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integration by Parts substitution

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    \int\arctan(4t)dt
    [/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]
    \int\arctan(4t)dt = t\arctan(4t) -4 \int \frac{t}{1+16t^2}dt
    [/tex]

    I'm stuck at this point. I think I need to make a substitution for the denominator, but I'm not sure how to go about doing so.
     
  2. jcsd
  3. Jan 30, 2010 #2

    rock.freak667

    User Avatar
    Homework Helper

    [tex]\int \frac{t}{1+16t^2}dt[/tex]


    f we let u=1+16t2 what is du equal to?
     
  4. Jan 30, 2010 #3
    It's good so far.

    You can substitute for [tex] V = {1+16t^2} [/tex], so [tex] dV = 32tdt[/tex].

    Then, you'll have [tex] \frac{1}{32} \int \frac{1}{V}dV[/tex].

    That equates to [tex] \frac {ln(|V|)}{32}[/tex].

    Multiply by -4, to obtain [tex] \frac {-ln(|V|)}{8}[/tex].

    Replace V and you're done.

    Good luck,

    Marc.
     
  5. Jan 30, 2010 #4
    Got it. Thank you both! Very helpful.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook