The integration about the function $\ln(\cos πx)$

Question

The problem is about how to integrate the function

Here is what I did

The stuck is that why we can get this

Answer 1

Let $I=\int_0^{1/2}\ln\cos\pi x dx=\pi^{-1}\int_0^{\pi/2}\ln\cos y dy$ so $I=\pi^{-1}\int_0^{\pi/2}\ln\sin y dy$ and $$2\pi I =\int_0^{\pi/2}\ln(\sin y\cos y) dy=\int_0^{\pi/2}\ln\sin 2y dy-\frac{\pi }{2}\ln 2=\frac{1}{2}\int_0^{\pi}\ln\sin x dx-\frac{\pi }{2}\ln 2.$$ Hence $$2\pi I=\int_0^{\pi/2}\ln\sin x dx-\frac{\pi }{2}\ln 2=\pi I -\frac{\pi }{2}\ln 2,$$ giving $I=-\frac{1}{2}\ln 2$ as you obtained.