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Integral Trig Substitution Question

  1. Oct 26, 2014 #1
    I just have a few questions. When using a trig substitution does it have to be under a radical ?

    eg, suppose I wanted to integrate (x2)/(x2-9), I used a trig substitution of x = 3sec(t) and got the wrong answer and so apparently I had to use partial fractions
     
  2. jcsd
  3. Oct 26, 2014 #2
    In the following integral,

    8281587fb9879173e9f84c4aa8e7d423.png

    Try using the following substitution:

    1e209acbb0838b8008b7b34fb7c8ebeb.png

    What do we have?
     
  4. Oct 26, 2014 #3

    Mark44

    Staff: Mentor

    No.

    Then I'm pretty sure you made a mistake. If you use partial fractions, you need to rewrite the integrand as a proper rational expression, which you can do by polynomial long division. (There's another way, as well.)
    You can also do this by trig substitution. Without seeing your work, I can't say why you got a wrong answer.
     
  5. Oct 26, 2014 #4
    Ok, I'm going to skip some of the substituting parts to this,

    3*integral sec3t/tan(t) dt
    = (1+tan2t)/(tan(t)) * sec(t) dt

    Integrating that, I get

    ln(sin(t)) + sec(t)

    Is it correct so far ?, I know the last thing I must do is sub the x values, but I just want to make sure if this part is correct
     
  6. Oct 26, 2014 #5

    Mark44

    Staff: Mentor

    This part looks OK. I'll check the following work in a bit.
     
  7. Oct 26, 2014 #6

    Mark44

    Staff: Mentor

    You lost the factor of 3, and have omitted the integration sign.
    BTW, it's easy enough to do in LaTeX - # # \int f(x)dx # # (omit the spaces between the pound signs).
    Nope. The factor of 3 is still missing on both terms. One of them should be 3 sec(t), but the other is wrong.
    I suspect that you did this:
    $$\int \frac{dt}{sin(t)} = ln(sin(t))$$
    ... which is incorrect. This is a somewhat tricky integral. Rather than deriving it here, I would advise just looking it up in a table of integrals.
     
  8. Oct 27, 2014 #7
    My integration was wrong. Turns out I had to use integration by parts for that trig integral with the sec^3 and tan and it was very messy. I did the same problem using partial fractions and it was much easier to integrate

    Thanks for the help, guys :)
     
  9. Oct 27, 2014 #8

    Mark44

    Staff: Mentor

    You don't need integration by parts. Starting from what you had in post #4
    In these integrals, sec(t) = x/3.
    $$\int \frac{3sec^3(t)}{tan(t)}dt$$
    $$=3\int \frac{dt}{cos^2(t) sin(t)}$$
    $$= 3\int sec^2(t)csc(t)dt$$
    $$=3\int \frac{sec(t)(tan^2(t) + 1)dt}{tan(t)}$$
    $$=3\int sec(t)tan(t)dt + 3 \int \frac{sec(t)dt}{tan(t)}$$
    The first integral is very easy, and the second integral simplifies to ##3\int csc(t)dt## which isn't too bad if you look it up in a table.

    After undoing the substitution, I get the same result as I got using partial fractions, after a bit of manipulation.
     
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