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efekwulsemmay
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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:
\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}} = 2
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.
\int^{a}_{-a} x dx = 2\cdot\int^{a}_{0} x dx
Let u = sin(t) + 2, du = cos(t)dt
So we start by saying:
\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}}
We can use the above property of integration to change this to:
2\cdot\int^{\frac{\pi}{2}}_{0} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}}
We then use u substitution thus we can say:
x=0 \rightarrow u=2
x= \frac{\pi}{2} \rightarrow u=3
so we get:
2\cdot\int^{3}_{2} \frac{3du}{u^{2}}
We can shove the 3 out front and then integrate the resulting \frac{du}{u^{2}}
Thus we get:
6\cdot\frac{-1}{u} Evaluated from 2 to 3.
This goes to:
6\cdot\left(\frac{-1}{3}-\frac{-1}{2}\right)
Which in turn goes to:
6\cdot\frac{1}{6} = 1
I don't know what I am doing wrong. Please help.
Homework Statement
The integral is:
\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}} = 2
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
\int^{a}_{-a} x dx = 2\cdot\int^{a}_{0} x dx
Let u = sin(t) + 2, du = cos(t)dt
The Attempt at a Solution
So we start by saying:
\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}}
We can use the above property of integration to change this to:
2\cdot\int^{\frac{\pi}{2}}_{0} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}}
We then use u substitution thus we can say:
x=0 \rightarrow u=2
x= \frac{\pi}{2} \rightarrow u=3
so we get:
2\cdot\int^{3}_{2} \frac{3du}{u^{2}}
We can shove the 3 out front and then integrate the resulting \frac{du}{u^{2}}
Thus we get:
6\cdot\frac{-1}{u} Evaluated from 2 to 3.
This goes to:
6\cdot\left(\frac{-1}{3}-\frac{-1}{2}\right)
Which in turn goes to:
6\cdot\frac{1}{6} = 1
I don't know what I am doing wrong. Please help.
G**d*** m*****f****** piece of s*** integral... grrrr. 
