# How to integrate the following

## Homework Statement

∫$\frac{sin\Theta}{1+sin^2\Theta}$ d$\Theta$ on [0,$\frac{\pi}{2}$]

## Homework Equations

I substituted:
u=cos$\Theta$
du=-sin$\Theta$

## 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.

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

You can use partial fractions or a trig substitution.

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 :)

HallsofIvy
Homework Helper

## Homework Statement

∫$\frac{sin\Theta}{1+sin^2\Theta}$ d$\Theta$ on [0,$\frac{\pi}{2}$]

## Homework Equations

I substituted:
u=cos$\Theta$
du=-sin$\Theta$

## The Attempt at a Solution

= ∫ $\frac{sin\Theta}{1+1-cos^2\Theta}$

= ∫ $\frac{sin\Theta}{2-cos\Theta}$

= ∫ $\frac{sin\Theta}{2-u^2}$
You've dropped the "$d\Theta$" in each of those- don't do that!

= $\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}$ <--- I'm pretty sure I shouldn't have done this.
You need, of course, "du" and you have the sign wrong in the denominator.

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

Any tips greatly appreciated.

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

$= \int\frac{\sin\theta}{1+1-\cos^2\theta}$
Again, don't forget the dθ!
$= \int \frac{\sin\theta}{2-\cos^2 \theta}$
$= \int \frac{\sin\theta}{2-u^2}$
Substitute in -du for sin θ dθ:
$$= \int \frac{-du}{2-u^2}$$
Can you take it from there?

EDIT: Beaten to it.