A calculus problem by Aly Ahmed

$\begin{aligned} \int_{0}^{\pi}\sin (x) \ln\left(\cot \frac{x}{2}\right)\mathrm{d}x &=\int_{0}^{\pi}\sin(\pi-x)\ln \left(\cot \left(\frac{\pi}{2}-\frac{x}{2}\right)\right) \mathrm{d}x \hspace{35pt}\text{using the property} \color{#3D99F6}{\int_{a}^{b}f(x)~\mathrm{d}x=\int_{a}^{b}f(a+b-x)~\mathrm{d}x} \\ &=\int_{0}^{\pi}\sin (x) \ln\left(\tan \frac{x}{2}\right)\mathrm{d}x \\ &=-\int_{0}^{\pi}\sin( x) \ln\left(\cot \frac{x}{2}\right)\mathrm{d}x \\ &=\boxed{0}. \end{aligned}$

Note that ln(cot(x/2)) = ln(cos(x/2)) - ln(sin(x/2)). Therefore, with a substitution of z = x/2 and 2dz = dx and the identity sin(2 z) = 2 sin(z)*cos(z), the integral becomes:

integral(ln(cos(z)) * 2 * sin(z) * cos(z), {z, 0, Pi/2}) - integral(ln(sin(z)) 2 sin(z)*cos(z), {z, 0, Pi/2})

In the first of these integrals, make the substitution u = cos(z), sin(z) dz = -du. In the second integral, make the substitution u = sin(z), du = cos(z) dz.

Upon doing this, we notice that the two integrals are additive inverses of one another, so that they sum to zero.