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

prove that  ∫_0 ^(π/2) ln(1−(1/2)sin2x)dx

$${prove}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}\right){dx} \\ $$

Commented by MJS_new last updated on 01/Dec/20

prove what?

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

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

∫_0 ^1 log(1−t(√(1−t^2 )))        =∫_0 ^1 ((log(1−t(√(1−t^2 ))))/( (√(1−t^2 ))))dt  =−Σ_(n=1) ^∞ (1/n)∫_0 ^1 t^n (1−t^2 )^((n−1)/2) dt       t^2 =u⇒2t=(du/dt)  =−(1/2)Σ_(n=1) ^∞ (1/n)∫_0 ^1 u^((n/2)−(1/2)) (1−u)^((n−1)/2) du  =−(1/2)Σ_(n=1) ^∞ ((Γ^2 (((n+1)/2)))/(nΓ(n+1)))    =−(1/2)Σ_(n=1) ^∞ ((Γ^2 (((n+1)/2)))/(n^2 Γ(n)))=−0.6403

$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{1}−{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right)\:\:\:\:\:\: \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{log}\left(\mathrm{1}−{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt} \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {dt}\:\:\:\:\:\:\:{t}^{\mathrm{2}} ={u}\Rightarrow\mathrm{2}{t}=\frac{{du}}{{dt}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{{n}\Gamma\left({n}+\mathrm{1}\right)}\:\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\Gamma^{\mathrm{2}} \left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} \Gamma\left({n}\right)}=−\mathrm{0}.\mathrm{6403} \\ $$

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