Showing that $\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$

Question

Wolfram Alpha tells me that $$\int_0^{\pi/4} \frac{1-\cos{16x}}{\sin{2x}}\,\mathrm{d}x=\frac{176}{105}$$

What are some quick/elegant ways of proving this?

Answer 1

$$1-\cos(16x) = 2\sin^2(8x) = 8 \sin^2(4x) \cos^2(4x) = 32 \sin^2(2x) \cos^2(2x) \cos^2(4x)$$ Hence, $$I = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) \cos^2(4x) dx = \int_0^{\pi/4}32 \sin(2x) \cos^2(2x) (2\cos^2(2x)-1)^2 dx$$ Now let $\cos(2x) = t$ and compute.