How to integrate the following

[dashhh]

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]

You can use partial fractions or a trig substitution.

 

[dashhh]

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 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]

(The LaTeX was a little messy, so I cleaned them up for you.)

I substituted:
$u = \cos \theta$
$du = -\sin \theta \;$dθ

Don't forget the dθ. Also, from here, move the negative sign:
$-du = \sin \theta \; d\theta$

$= \int\frac{\sin\theta}{1+1-\cos^2\theta}$dθ

Again, don't forget the dθ!

$= \int \frac{\sin\theta}{2-\cos^2 \theta}$dθ
$= \int \frac{\sin\theta}{2-u^2}$dθ

Substitute in -du for sin θ dθ:
$$= \int \frac{-du}{2-u^2}$$
Can you take it from there?

EDIT: Beaten to it.