urgent integration help!!!

[sonia11]

I really need an urgent help on this question:

Integration of ( sin[x] ) ^ 3 / ( 1 + cos[x] ) ^ 4 dx from limit pi/4(lower limit) to pi/2 (upper limit).
I am really confused if we should take 1+cos[x] as the 'u' (substitution) or not because that completely changes the limit where upper limit becomes smaller than the lower limit.
Also when I am taking 1 - cos[x] (from sin^2 formula) as 'u' (substitution), then I am getting a very confusing value to integrate.
Any help for the problem will be highly highly appreciated.
Thanks in advance.

 

[BigGlenntheHeavy]

$\displaystyle Use \ u \ = \ \frac{sin(x)}{1+cos(x)} \ = \ tan(x/2). \ This \ yields,$

$\displaystyle cos(x) \ = \ \frac{1-u^2}{1+u^2}, \ sin(x) \ = \ \frac{2u}{1+u^2}, \ and$

$\displaystyle dx \ = \ \frac{2du}{1+u^2}.$

$\displaystyle This \ will \ do \ it.$