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Troublesome integral problem

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data
    evaluate the integral from 0 to e^x of sin(tan^-1(e^(sqrt (x)))dx


    2. Relevant equations
    all trig function such as sin=opp/hypo, cos=adj/hypo, tan oppo/adj


    3. The attempt at a solution
    I believe the integral can be written as the integral from 0 to e^x of (e^sqrt(x))/(sqrt(1+e^(2sqrt(x))).

    Anyone know how to solve this problem?
     
  2. jcsd
  3. Nov 25, 2013 #2

    Mark44

    Staff: Mentor

    Start by sketching a right triangle and label the sides and an acute angle θ so that tan(θ) = e√x. Find sin(θ).
     
  4. Nov 25, 2013 #3
    using that i can get integral from 0 to e^x of (e^sqrt(x))/(sqrt(1+e^(2sqrt(x)))

    how should I continue
     
  5. Nov 25, 2013 #4
    using that I can find integral from 0 to e^x of (e^sqrt(x))/(sqrt(1+e^(2sqrt(x)))

    how should I continue
     
  6. Nov 25, 2013 #5

    Mark44

    Staff: Mentor

    Where is dx? If you are in the habit of ignoring it and omitting it, you will come to grief, and things will become very difficult.

    Here's the integral that you are reporting in post #4:
    $$ \int_0^{e^x} \frac{e^{\sqrt{x}}~dx}{\sqrt{1 + e^{2\sqrt{x}}}}$$

    It's not a very good practice to have the same variable in one or both limits of integration as in the "dummy" variable of the integrand. Let's change that.
    $$ \int_{t = 0}^{t = e^x} \frac{e^{\sqrt{t}}~dt}{\sqrt{1 + e^{2\sqrt{t}}}}$$

    That looks better.

    To make forward probress, an ordinary substitution might be called for here.
     
  7. Nov 25, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I don't think any simple substituition will deal with that, it's a very nasty integral. I think you need nonelementary functions to deal with it, like polylogarithms. I suspect rich_machine isn't telling us the whole question. Do you want to find the derivative of that integral? Finding the derivative of an integral doesn't require finding the integral first.
     
    Last edited: Nov 25, 2013
  8. Nov 26, 2013 #7

    Mark44

    Staff: Mentor

    Mostly, I was looking for directions to go, and it was bothersome that with the substitution I was heading towards, there wasn't anything to make up du.
    This thought did cross my mind.
     
  9. Nov 26, 2013 #8
    Thank you dick and mark. I believe my calc professor has assigned a problem well beyond the scope of the class. Most likely he created the problem himself at a moments notice and did not realize the difficulty of the integral.
     
  10. Nov 26, 2013 #9

    haruspex

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    Gold Member
    2016 Award

    I assume the tan^-1 means arctan, not cotan.

    My inclination is to get rid of the arctan first, setting it to theta:
    ##\int\sin(\theta).dx## where ##x = ln(2\tan(\theta))##, so ##dx = \frac 1{\sin(\theta)\cos(\theta)}dθ##.
    Looks easy from there.
     
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