# Integration problem

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:
$$\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{3\cos(t)dt}{\left(2+\sin(t)\right)^{2}} = 2$$

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:
$$6\cdot\frac{1}{6} = 1$$

## Homework Equations

$$\int^{a}_{-a} x dx = 2\cdot\int^{a}_{0} x dx$$

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

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. 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