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

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 $\ln 3$.
