# How to integrate the following

1. Jul 19, 2012

### dashhh

1. The problem statement, all variables and given/known data
∫$\frac{sin\Theta}{1+sin^2\Theta}$ d$\Theta$ on [0,$\frac{\pi}{2}$]

2. Relevant equations
I substituted:
u=cos$\Theta$
du=-sin$\Theta$

3. The attempt at a solution
= ∫ $\frac{sin\Theta}{1+1-cos^2\Theta}$

= ∫ $\frac{sin\Theta}{2-cos\Theta}$

= ∫ $\frac{sin\Theta}{2-u^2}$

= $\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}$ <--- I'm pretty sure I shouldn't have done this.

And that's as far as I've gotten.

Any tips greatly appreciated.

2. Jul 19, 2012

### vela

Staff Emeritus
You can use partial fractions or a trig substitution.

3. Jul 19, 2012

### dashhh

We haven't done trig substitution, I think partial fractions is covered in the next section of my notes though so I'll check it out. Thanks :)

4. Jul 19, 2012

### HallsofIvy

Staff Emeritus
You've dropped the "$d\Theta$" in each of those- don't do that!

You need, of course, "du" and you have the sign wrong in the denominator.

5. Jul 19, 2012

### eumyang

(The LaTeX was a little messy, so I cleaned them up for you.)
Don't forget the dθ. Also, from here, move the negative sign:
$-du = \sin \theta \; d\theta$

Again, don't forget the dθ!
Substitute in -du for sin θ dθ:
$$= \int \frac{-du}{2-u^2}$$
Can you take it from there?

EDIT: Beaten to it.