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## 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.