How to integrate the following

[dashhh]

Homework Statement

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

Homework Equations

I substituted:
u=cos[itex]\Theta[/itex]
du=-sin[itex]\Theta[/itex]

The Attempt at a Solution

= ∫ [itex]\frac{sin\Theta}{1+1-cos^2\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-cos\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-u^2}[/itex]

= [itex]\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}[/itex] <--- 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

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

Homework Equations

I substituted:
u=cos[itex]\Theta[/itex]
du=-sin[itex]\Theta[/itex]

The Attempt at a Solution

= ∫ [itex]\frac{sin\Theta}{1+1-cos^2\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-cos\Theta}[/itex]

= ∫ [itex]\frac{sin\Theta}{2-u^2}[/itex]

You've dropped the "[itex]d\Theta[/itex]" in each of those- don't do that!

= [itex]\frac{1}{(u+\sqrt{2})(u-\sqrt{2})}[/itex] <--- 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:
[itex]u = \cos \theta[/itex]
[itex]du = -\sin \theta \;[/itex]dθ

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

[itex]= \int\frac{\sin\theta}{1+1-\cos^2\theta}[/itex]dθ

Again, don't forget the dθ!

[itex]= \int \frac{\sin\theta}{2-\cos^2 \theta}[/itex]dθ
[itex]= \int \frac{\sin\theta}{2-u^2}[/itex]dθ

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

EDIT: Beaten to it.