Question Number 77160 by Chi Mes Try last updated on 03/Jan/20
Commented by Chi Mes Try last updated on 03/Jan/20
$${please}\:{help}\:{out} \\ $$
Answered by mind is power last updated on 05/Jan/20
![∫_0 ^π ((xsin(x))/(1+cos^2 (x)))=∫_0 ^π (((π−x)sin(x))/(1+cos^2 (x)))dx ⇒2∫((xsin(x))/(1+cos^2 (x)))dx=π∫_0 ^π ((sin(x))/(1+cos^2 (x)))dx=[−πarctan(cos(x))]_0 ^π =(π^2 /2)⇒∫_0 ^π ((xsin(x))/(1+cos^2 (x)))dx=(π^2 /4) (k((1/(√2))))^2 =(∫_0 ^(π/2) (((√2)dx)/(√(2−sin^2 (x)))))^2 =2(∫_0 ^(π/2) (dx/(√(2−sin^2 (x)))))^2 Γ((1/4)).Γ(1−(1/4))=(π/(sin((π/4))))=π(√2) ((Γ(x).Γ(y))/(Γ(x+y)))=∫_0 ^1 t^(x−1) (1−t)^(y−1) dt 2x=y=(1/2)⇒ ((Γ((1/4)).Γ((1/2)))/(Γ((3/4))))=∫_0 ^1 t^((−3)/4) .(1−t)^(−(1/2)) dt t=u^4 ⇒du=(1/4)t^(−(3/4)) dt ⇒((Γ((1/4)).(√π))/(Γ((3/4))))=4∫_0 ^1 (1−u^4 )^(−(1/4)) du Γ((1/4)).Γ((3/4))=π(√2) ⇒((Γ^2 ((1/4)))/(√(2π)))=4∫_0 ^1 (dx/(√(1−x^4 ))) k((1/(√2)))^2 =2(∫_0 ^(π/2) (dx/(√(2−sin^2 (x)))))^2 =2(∫_0 ^(π/2) (dx/(√(1+cos^2 (x)))))^2 =2(∫_0 ^1 (dx/(√(1−x^4 ))))^2 =k((1/(√2)))^2 ⇒((Γ^2 ((1/4)))/(4(√(2π))))=∫_0 ^1 (dx/(√(1−x^4 ))) ⇒2(∫_0 ^1 ((dx/(√(1−x^4 )))))^2 =k^2 ((1/(√2)))=((Γ^4 ((1/4)))/(16π))=((Γ^4 ((1/4)))/(4π)) ((k^2 ((1/(√2))))/(∫_0 ^π ((xsin(x))/(1+cos^2 (x)))dx))=((Γ^4 ((1/4)))/((16π)/(π^2 /4)))=((Γ^4 ((1/4)))/(4π^3 ))](Q77349.png)
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−\mathrm{x}\right)\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx} \\ $$$$\Rightarrow\mathrm{2}\int\frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx}=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx}=\left[−\pi\mathrm{arctan}\left(\mathrm{cos}\left(\mathrm{x}\right)\right)\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\left(\mathrm{k}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right)^{\mathrm{2}} =\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\sqrt{\mathrm{2}}\mathrm{dx}}{\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}\right)^{\mathrm{2}} =\mathrm{2}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}\right)^{\mathrm{2}} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)}=\pi\sqrt{\mathrm{2}} \\ $$$$\frac{\Gamma\left(\mathrm{x}\right).\Gamma\left(\mathrm{y}\right)}{\Gamma\left(\mathrm{x}+\mathrm{y}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{x}−\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{y}−\mathrm{1}} \mathrm{dt} \\ $$$$\mathrm{2x}=\mathrm{y}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow \\ $$$$\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\frac{−\mathrm{3}}{\mathrm{4}}} .\left(\mathrm{1}−\mathrm{t}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dt} \\ $$$$\mathrm{t}=\mathrm{u}^{\mathrm{4}} \Rightarrow\mathrm{du}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \mathrm{dt} \\ $$$$\Rightarrow\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\sqrt{\pi}}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{u}^{\mathrm{4}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} \mathrm{du} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\pi\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\sqrt{\mathrm{2}\pi}}=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }} \\ $$$$\mathrm{k}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\mathrm{2}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\right)^{\mathrm{2}} =\mathrm{k}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \Rightarrow\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\sqrt{\mathrm{2}\pi}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }} \\ $$$$\Rightarrow\mathrm{2}\left(\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{dx}}{\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\right)\right)^{\mathrm{2}} =\mathrm{k}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{16}\pi}=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\pi} \\ $$$$\frac{\mathrm{k}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)}{\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)}\mathrm{dx}}=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\frac{\mathrm{16}\pi}{\frac{\pi^{\mathrm{2}} }{\mathrm{4}}}}=\frac{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\pi^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$