# An integration of sorts.

• Sep 10th 2008, 01:21 AM
Showcase_22
An integration of sorts.
http://i116.photobucket.com/albums/o...sproblem18.jpg

That's the question I can't do. It would be so much easier if it was a cos squared!

I think I have to use a substitution but i'm not quite sure what it should be.(Doh)
• Sep 10th 2008, 01:27 AM
Chop Suey
Quote:

Originally Posted by Showcase_22
http://i116.photobucket.com/albums/o...sproblem18.jpg

That's the question I can't do. It would be so much easier if it was a cos squared!

I think I have to use a substitution but i'm not quite sure what it should be.(Doh)

Multiply by $\frac{\sqrt{1+\cos{\theta}}}{\sqrt{1+\cos{\theta}} }$

Can you do it now?
• Sep 10th 2008, 01:42 AM
Chop Suey
Or, you can rewrite $\sqrt{1-\cos{\theta}}$ as $\sqrt{2}\sin{\left(\frac{x}{2}\right)}$. I'll leave it to you to figure out why they're equivalent.
• Sep 10th 2008, 02:39 AM
Showcase_22
I can do it, and I have solved it.(Rofl)

I tried deriving the second identity myself but I couldn't figure that part out.

can you show me how it's done?
• Sep 10th 2008, 02:50 AM
Chop Suey
$\sqrt{1-\cos{\theta}}$

$= \displaystyle\sqrt{\underbrace{1-\cos^2{\frac{\theta}{2}}}_{\sin^2{\frac{\theta}{2} }}+\sin^2{\frac{\theta}{2}}}$

$= \sqrt{2\sin^2{\frac{\theta}{2}}}$

$= \sqrt{2}\sin{\frac{\theta}{2}}$
• Sep 10th 2008, 02:52 AM
Showcase_22
oh!! I don't know what I was doing. It's just a simple double angle formulae.

That's rather useful.(Clapping)