Question Number 124177 by mathdave last updated on 01/Dec/20
$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}\right){dx} \\ $$
Commented by MJS_new last updated on 01/Dec/20
$$\mathrm{prove}\:\mathrm{what}? \\ $$
Commented by Dwaipayan Shikari last updated on 01/Dec/20
$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}−{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right)\:\:\:\:\:\: \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {dt}\:\:\:\:\:\:\:{t}^{\mathrm{2}} ={u}\Rightarrow\mathrm{2}{t}=\frac{{du}}{{dt}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{{n}\Gamma\left({n}+\mathrm{1}\right)}\:\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} \Gamma\left({n}\right)}=−\mathrm{0}.\mathrm{6403} \\ $$