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

Homework Help: Integration problem

  1. Mar 19, 2010 #1
    I have no idea what I am doing wrong. I keep getting one when I should be getting two. It is part of a numerical integration problem. I've got the numerical integration part down which is ironic. The part I am having problems with is finding the actual value of the integral. I need this to find the error of the trapezoid and Simpson's estimations.

    1. The problem statement, all variables and given/known data
    The integral is:
    [tex]\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}} = 2[/tex]

    I know it equals 2 cause of the integrate function on my calculator. I am trying to figure out where I am going wrong with my algebra.

    2. Relevant equations
    [tex]\int^{a}_{-a} x dx = 2\cdot\int^{a}_{0} x dx[/tex]

    [tex]Let u = sin(t) + 2, du = cos(t)dt[/tex]

    3. The attempt at a solution
    So we start by saying:
    [tex]\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}} [/tex]

    We can use the above property of integration to change this to:
    [tex]2\cdot\int^{\frac{\pi}{2}}_{0} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}} [/tex]

    We then use u substitution thus we can say:
    [tex]x=0 \rightarrow u=2[/tex]

    [tex]x= \frac{\pi}{2} \rightarrow u=3[/tex]

    so we get:
    [tex]2\cdot\int^{3}_{2} \frac{3du}{u^{2}} [/tex]

    We can shove the 3 out front and then integrate the resulting [tex]\frac{du}{u^{2}}[/tex]
    Thus we get:
    [tex]6\cdot\frac{-1}{u}[/tex] Evaluated from 2 to 3.

    This goes to:
    [tex]6\cdot\left(\frac{-1}{3}-\frac{-1}{2}\right)[/tex]

    Which in turn goes to:
    [tex]6\cdot\frac{1}{6} = 1[/tex]

    I don't know what I am doing wrong. Please help.
     
  2. jcsd
  3. Mar 19, 2010 #2

    jav

    User Avatar

    This is true if the function is even. Try not using this when you solve for the integral. ie. use 1 and 3 as your new bounds for your u integral
     
  4. Mar 19, 2010 #3
    Oh bloody hell. :mad: G**d*** m*****f****** piece of s*** integral..... grrrr.

    Thank you for your help jav. I truly appreciate it. Now I must bang my head against a brick wall somemore :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook