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Prove that $$ \int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x(x+1)}dx=\frac{1}{2} (\ln 2)^2-\ln \pi \ln2 $$

I separated them $$ \int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx $$

For the former integral i tried to use differentiation under integration but got stuck and i have no idea about the latter one. Plz help!

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    $\begingroup$ Please indicate what you have tried and where you are stuck. This will allow others to better tailor their answers to your specific background and ability level. Also it will show others you are willing to learn and haven't just come here to get your homework done for you. $\endgroup$ – Ian Miller Dec 7 '15 at 16:36
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    $\begingroup$ where does this problem arise, have u any sources? $\endgroup$ – tired Dec 7 '15 at 17:47
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    $\begingroup$ @MartinNicholson Here. This Tanishq is a cheater $\endgroup$ – GohP.iHan Dec 10 '15 at 3:44
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    $\begingroup$ @tanishq y did u cheat man! U spoilt the contest! $\endgroup$ – Aditya Kumar Dec 10 '15 at 5:39
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    $\begingroup$ Do you like it if you posted a question in Brilliant, and someone else asked the same question in MSE to get the answer and then posted the exact same answer? At the very least, you can cite where you get the answer. $\endgroup$ – GohP.iHan Dec 10 '15 at 18:37
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\begin{align} &\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx=\\ &\int^{1}_{0} \frac{\ln\frac{\sin \pi x}{2\sin(\pi x/2)}}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx=\\ &\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx\right)=\\ &\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{2}_{1} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx\right)=\\ &\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{2}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx\right)=\\ &\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon/2} \frac{\ln \left(\sin \pi x\right)}{x}dx\right)=\\ &\lim_{\varepsilon\to 0}\left(-\ln 2\int^{1}_{\varepsilon}\frac{dx}{x}-\int^{\varepsilon}_{\varepsilon/2} \frac{\ln \left(\sin \pi x\right)}{x}dx \right)=\\ &\lim_{\varepsilon\to 0}\left(-\ln 2\int^{1}_{\varepsilon}\frac{dx}{x}-\int^{\varepsilon}_{\varepsilon/2} \frac{\ln \left(\pi x\right)}{x}dx \right)=\\ &\lim_{\varepsilon\to 0}\left(\ln 2\ln\varepsilon-\ln\pi\ln 2-\frac{1}{2}\left[(\ln\varepsilon)^2-(\ln\varepsilon/2)^2\right]\right)=\frac{1}{2} (\ln 2)^2-\ln \pi \ln2 \end{align}

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  • $\begingroup$ Very clever, (+1) $\endgroup$ – tired Dec 8 '15 at 12:12

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