Question Number 205361 by Lambertician last updated on 18/Mar/24
Answered by Berbere last updated on 18/Mar/24
![=−4∫_0 ^1 ((xln(((1−x)/(1+x))))/( (√(1−x^2 ))))dx−∫_0 ^1 (1/( (√(x(1−x))))) =−4a−q q=∫_0 ^1 x^((1/2)−1) (1−x)^((1/2)−1) dx=β((1/2),(1/2))=((Γ^2 ((1/2)))/(Γ(1)))=π a=∫_0 ^1 ((xln(((1−x)/(1+x))))/( (√(1−x^2 ))))dx^(IBP) =lim_(x→1) [−(√(1−x^2 ))ln(((1−x)/(1+x)))]_0 ^x +∫_0 ^1 (√(1−x^2 )).(((−2)/((1+x)^2 ))/((1−x)/(1+x)))dx =∫_0 ^1 ((−2)/( (√(1−x^2 ))))=−2[sin^(−1) (x)]_0 ^1 =−π −4.−π−π=3π](https://www.tinkutara.com/question/Q205362.png)
$$=−\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{x}}\left(\mathrm{1}−\boldsymbol{{x}}\right)}} \\ $$$$=−\mathrm{4}{a}−{q} \\ $$$${q}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {dx}=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{1}\right)}=\pi \\ $$$${a}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{d}\overset{{IBP}} {{x}}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)\right]_{\mathrm{0}} ^{{x}} +\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{2}} }.\frac{\frac{−\mathrm{2}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }}{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=−\mathrm{2}\left[\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =−\pi \\ $$$$−\mathrm{4}.−\pi−\pi=\mathrm{3}\pi \\ $$