Solving Integration Problem: I Have No Idea What I'm Doing Wrong

  • Thread starter efekwulsemmay
  • Start date
  • Tags
    Integration
In summary, the person is having trouble with the integral and is seeking guidance from a friend. They are using a property of integration that is true for even functions, but is not applicable in this situation.
  • #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.

Homework Statement


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.

Homework 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]

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.
 
Physics news on Phys.org
  • #2
efekwulsemmay said:

Homework Equations


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

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
 
  • #3
jav said:
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

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:
 

Suggested for: Solving Integration Problem: I Have No Idea What I'm Doing Wrong

Back
Top