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Trig substitution: integrate sqrt(16+x^2) over x

  1. Jan 31, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    set x=4tant
    dx=4sec[itex]^{2}[/itex]t dt

    so after plugging in and using a quick trig identity I get:

    [itex]\int\frac{16(sec^{2}t)*4sec^{2}t dt}{4tant}[/itex]

    Then after a quick cleanup:

    16*[itex]\int tan^{-1}t*sec^{4}t *dt[/itex]

    I think I made a mistake, but if not, what next?
  2. jcsd
  3. Jan 31, 2012 #2


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    Best not to use ##\tan^{-1}## for ##1/\tan## since that may be confused with arctan. But write ##\sec^4(x)= \sec^2 x(1 + \tan^2 x)## and try a u - substitution ##u=\tan x##.
  4. Jan 31, 2012 #3
    I think you've made a mistake, where has the square root gone?
  5. Jan 31, 2012 #4
    I definitely made a mistake. I forgot the square root sign, which changes everything. Thanks!
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