Show that $\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}?$ 
$$\int_{0}^{\pi}{\sin^5(x)\over 1+\cos^3(x)}\mathrm dx =\ln{3}\tag1$$

How can we show that $(1)=\ln{3}$
$u=\sin^3{x}$ then $du=3\sin^2{x}\cos{x}dx$
$${1\over 3}\int{u\mathrm du\over \cos{x}+\cos^4{x}}\tag2$$
This is not a good substitution
 A: $$\int_0^\pi \frac{\sin^5 x}{1+\cos^3 x}dx$$
$$=\int_0^\pi \frac{-(1-\cos^2 x)^2}{1+\cos^3 x}\cdot -\sin x\space dx$$
Now let $u=\cos x$. Then we have
$$=\int_1^{-1} \frac{-(1-u^2)^2}{1+u^3}\cdot du$$
$$=\int_{-1}^{1} \frac{(1-u^2)^2}{1+u^3}\cdot du$$
Can you finish this?
A: HINT
Note that: $$sin^5(x)=\sin^2 x \sin^2 x \sin x=(1-cos^2 x)^2 \sin x$$
This suggest to set: $\cos x =u$
A: Try:
$$ u = \cos(x) \\ du = -\sin(x)dx
$$
Then
$$\frac{sin^5(x)}{1+\cos^3(x)}dx  = -\frac{(1-u^2)^2}{1+u^3}du$$
A: Without using integration by substitution or by parts:
Trigonometric algebra
We use the identity that $(1-\cos x)(1+ \cos^3x)=(1-\cos^2x)(1-\cos x+\cos^2x)$ so $$\frac{\sin^2x}{1+\cos^3x}=\frac{1-\cos x}{1-\cos x + \cos^2x} \implies \frac{\sin^5x}{1+\cos^3x}=\frac{\sin^3x(1-\cos x)}{1-\cos x + \cos^2x}$$ Now RHS is equivalent to $$\frac{\sin x(1-\cos^2x)(1-\cos x)}{1-\cos x + \cos^2x}=\frac{\sin x-\sin x \cos x+\sin x \cos^2x(\cos x-1)}{1-\cos x + \cos^2x}$$ or $$\frac{\sin x(1-2\cos x)+\sin x \cos x(1-\cos x + \cos^2x)}{1-\cos x + \cos^2x}=\frac{2\sin x(1-\cos^2x)}{2-2\cos x + 2\cos^2x}+\sin x \cos x$$ We do this so that we can replace the square term with $\cos 2x$. Continuing, we have $$\frac{2\sin x - 4 \sin x \cos x}{3-2\cos x + \cos 2x}+\frac12 \sin 2x=\frac{-2(-\sin x)+2(-\sin 2x)}{3-2\cos x + \cos 2x}-\frac12(-\sin 2x)$$
Notice that the numerator is the derivative of the denominator! Thus the integral is $$\int \frac{\sin^5x}{1+\cos^3x} dx=\ln|3-2\cos x + \cos 2x|-\frac14\cos 2x + c$$ where $c$ is a constant. 
Substituting the upper and lower limits of $\pi$ and $-\pi$ respectively yields $\ln 3$.
A: Do you know about the Bioche's rules for rational functions of $\cos,\sin$.
Wiki : Règles de Bioche
In our case $w(x)=\dfrac{\sin(x)^5}{1+\cos(x)^3}\mathop{dx}$ and we can see that $w(-x)=w(x)$ 
Thus a substitution $\boxed{u=\cos(x)}$ is appropriate.
After some calculation you get $\displaystyle I=\int_{-1}^1 \left(u-\dfrac{2u-1}{u^2-u+1}\right)\mathop{du}$ with a nice $\frac {f'}{f}$ appearance.
A: $$\int_{0}^{\pi}\frac{\sin^5(x)}{1+\cos^3(x)}\,dx = \int_{0}^{\pi/2}\frac{\sin^5(x)}{1+\cos^3(x)}+\frac{\sin^5(x)}{1-\cos^3(x)}\,dx $$
equals:
$$ \int_{0}^{\pi/2}\frac{2\sin^5(x)}{1-\cos(x)^6}\,dx = \int_{0}^{\pi/2}\frac{2\cos(x)^5}{1-\sin(x)^6}\,dx $$
or, by setting $\sin(x)=u$,
$$ \int_{0}^{1}\frac{2\sqrt{1-u^2}^4}{1-u^6}\,du=2\int_{0}^{1}\frac{1-u^2}{1+u^2+u^4}\,du $$
which is an elementary integral (by partial fraction decomposition).
