The problem is about how to integrate the function
Here is what I did
The stuck is that why we can get this 
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- 3$\begingroup$ This isn't a physics question. $\endgroup$Chris– Chris2018-02-09 05:56:17 +00:00Commented Feb 9, 2018 at 5:56
- $\begingroup$ We'd prefer you use Mathjax for equations, not images. It's the site standard. $\endgroup$StephenG - Help Ukraine– StephenG - Help Ukraine2018-02-09 06:29:27 +00:00Commented Feb 9, 2018 at 6:29
- $\begingroup$ What is $d^2 x$? See math.stackexchange.com/questions/341643/… and math.stackexchange.com/questions/333941/… $\endgroup$FDP– FDP2018-02-09 11:43:07 +00:00Commented Feb 9, 2018 at 11:43
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1 Answer
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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.
