A calculus problem by Aly Ahmed

$\begin{aligned} I & = \int_0^\pi \cot x \sin 4x\ dx \\ & = \int_0^\pi 2 \cot x \sin 2x \cos 2x \ dx \\ & = \int_0^\pi \frac {4\cos x \sin x \cos x \cos 2x} {\sin x} \ dx \\ & = \int_0^\pi 4 \cos^2 x \cos 2x \ dx \\ & = \int_0^\pi 2(1+\cos 2 x) \cos 2x \ dx \\ & = \int_0^\pi(2\cos 2 x +2 \cos^2 2x) \ dx \\ & = \int_0^\pi(2\cos 2 x +1+\cos 4x) \ dx \\ &=\sin 2x + x + \frac {\sin 4x}4\ \bigg|_0^\pi \\ & = \pi \approx \boxed{3.14} \end{aligned}$

$\sin (4x)=4\cos^3 x\sin x-4\cos x\sin^3 x$

So, the integrand is $4\cos^4 x-4\cos^2 x\sin^2 x=4\cos^2 x(\cos^2 x-\sin^2 x)$

The indefinite integral is $\dfrac {\sin (4x)+4\sin (2x)+4x}{4}+C$

Hence, the value of the definite integral is $π\approx \boxed {3.14}$ .