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Question Number 124177 by mathdave last updated on 01/Dec/20

$$provethat$$$${\int }_0^{\frac{\pi }{2}}\ln (1-\frac{1}{2}\sin 2x)dx$$

Commented by MJS_new last updated on 01/Dec/20

$$\mathrm{prove}\mathrm{what}?$$

Commented by Dwaipayan Shikari last updated on 01/Dec/20

$${\int }_0^{1}log(1-t\sqrt{1-t^2})$$$$={\int }_0^1\frac{log(1-t\sqrt{1-t^2})}{\sqrt{1-t^2}}dt$$$$=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{t}^n{(1-t^2)}^{\frac{n-1}{2}}dt{t}^2=u\Rightarrow 2t=\frac{du}{dt}$$$$=-\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n}{\int }_0^1{u}^{\frac{n}{2}-\frac{1}{2}}{(1-u)}^{\frac{n-1}{2}}du$$$$=-\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^2\left(\frac{n+1}{2}\right)}{n\mathrm{\Gamma }(n+1)}$$$$=-\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{\Gamma }}^2\left(\frac{n+1}{2}\right)}{n^2\mathrm{\Gamma }\left(n\right)}=-0.6403$$

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