- #1
- 54
- 0
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.
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.
[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]
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.
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.