Integrating trigonometric quotients This is my final question:


*Consider the function $ g(x) = \frac{\sin^3 x}{
1 + \cos x}$.
Write $g(x) = f(u, v)$, where $u = \sin x$ and $v = \cos x$.
Consulting Section 18.16 on p.299 of Ostaszewski’s book if necessary, evaluate $\int_0^{π/2}\frac{\sin^3 x}{
1 + \cos x}dx$



I got the answer using this method: 
$\sin^3x=\sin^2x\times \sin x=(1-\cos^2x)\times\sin x=(1+\cos x)(1-\cos x)\times \sin x$
And then cancelling out the $1+\cos x$ term. Is this the best way to tackle this question or was I being expected to use some other method? I don't have the book that the question references so I wasn't sure. I feel like I was supposed to do something with the fact that $g(x)=\frac{u^3}{1+v}$.
 A: $I=\int_0^{\pi/2}\dfrac{\sin^3(x)}{1+\cos(x)}dx=\int_0^{\pi/2}\dfrac{\sin(x)(1-\cos^2(x))}{1+\cos(x)}dx$
Substitute $u=\cos(x)$
$I =-\int_1^0\dfrac{1-u^2}{1+u}du=\int_0^1(1-u)du=\dfrac{1}2$ 
A: Ok I bought the book ... £62 :'( lol
Basically the book states a set of substitutions to use according to the nature of the function $f(\sin x, \cos x)$
In this case it states that if we consider that $f(\sin x, \cos x) = \frac{\sin^3 x}{1+\cos x}$ is an odd function with respect to $\sin$, i.e. $f(\sin x, \cos x) = -f(-\sin x,\cos x)$ then $w=\ cos x$ is a suitable substitution.
$w = \cos x \rightarrow \frac{dw}{dx}=-\sin x \rightarrow dx = \frac{-dw}{\sin x}$
$\int_{x=0}^{\pi / 2} \frac{\sin^3x}{1+\cos x}dx$
$=\int_{w=1}^{0} \frac{\sin^3x}{1+ w}\times \frac{-1}{sin x}dw$
$=\int_1^0 \frac{-\sin^2x}{1+w}dw$
$=-\int_1^0 \frac{1-\cos^2x}{1+w}dw$
$=-\int_1^0 \frac{1 - w^2}{1+w}dw$
$=-\int_1^0 \frac{(1+w)(1-w)}{1+w}dw$
$=-\int_1^0 (1-w)dw$
$=-[w-\frac{w^2}{2}]_1^0$
$=-[(0-\frac{0^2}{2})-(1-\frac{1^2}{2})]$
$=-[(0)-(1-\frac{1}{2})]$
$=-[(0)-(\frac{1}{2})]$
$=-[-\frac{1}{2}]$
$=\frac{1}{2}$
