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Question Number 118705 by Lordose last updated on 19/Oct/20

$$$$$$$$$$...⧫\mathrm{Advanced}\mathrm{Calculus}⧫...$$$$$$$$\mathrm{Evaluate}::$$$$$$$$\mathrm{\Omega }={\int }_0^1\frac{{\sin }^{-1}\left(\mathrm{x}\right)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$$$$$...\mathrm{♠}\mathbf{L}\mathit{\varphi }\mathbf{rD}\mathit{\varnothing }\mathbf{sE}\mathrm{♠}...$$$$$$$$...\mathrm{♣}\mathbf{GooD}\mathbf{LucK}\mathrm{♣}$$

Answered by mathdave last updated on 19/Oct/20

$$solution$$$$letI={\int }_0^1\frac{{\sin }^{-1}x}{1+x^2}dx(puty={\sin }^{-1}x)$$$$I={\int }_0^{\frac{\pi }{2}}\frac{y\cos y}{1+{\sin }^{2}y}dy\left(usingIBP\right)$$$$I={\left(y{\tan }^{-1}\left(\sin y\right)\right)}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\sin y\right)dy$$$$I=\frac{{\pi }^2}{8}-{\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\sin y\right)dy$$$$accordingtolegendarychi-function$$$${\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(r\sin \theta \right)d\theta =2{\chi }_2\left(\frac{\sqrt{1+r^2}-1}{r}\right)$$$$ifr=1then$$$${\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\sin y\right)dy=2{\chi }_2(\sqrt{2}-1)$$$$\because notethat{\chi }_v\left(z\right)=\frac{1}{2}({Li}_v\left(z\right)-{Li}_v(-z))$$$$then$$$${\chi }_2(\sqrt{2}-1)=\frac{1}{2}({Li}_2(\sqrt{2}-1)-{Li}_2(1-\sqrt{2}))$$$$\because {\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\sin y\right)dy=2\cdot \frac{1}{2}({Li}_2(\sqrt{2}-1)-{Li}_2(1-\sqrt{2}))$$$$\because I=\frac{{\pi }^2}{8}-({Li}_2(\sqrt{2}-1)-{Li}_2(1-\sqrt{2}))$$$$\because {\int }_0^1\frac{{\sin }^{-1}\left(x\right)}{1+x^2}dx=\frac{{\pi }^2}{8}-({Li}_2(\sqrt{2}-1)-{Li}_2(1-\sqrt{2})=0.38841$$$$OR$$$$legendarychi-functionof$$$$2{\chi }_2(\sqrt{2}-1)=2(\frac{{\pi }^2}{16}-\frac{{\ln }^2(\sqrt{2}+1)}{4})=\frac{{\pi }^2}{8}-\frac{{\ln }^2(\sqrt{2}+1)}{2}$$$$\because {\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\sin y\right)dy=(\frac{{\pi }^2}{8}-\frac{{\ln }^2(\sqrt{2}+1)}{2})$$$$\because I=\frac{{\pi }^2}{8}-(\frac{{\pi }^2}{8}-\frac{{\ln }^2(\sqrt{2}+1)}{2})=\frac{{\ln }^2(\sqrt{2}+1)}{2}$$$$\because {\int }_0^1\frac{{\sin }^{-1}\left(x\right)}{1+x^2}dx=\frac{{\ln }^2(\sqrt{2}+1)}{2}=0.38841$$$$thetwoanswerarecorrect$$$$bymathdave\left(19/10/2020\right)$$

Commented by Lordose last updated on 19/Oct/20

$$\mathrm{Nice}\mathrm{one}\mathrm{sir}$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by mindispower last updated on 19/Oct/20

$$x=sin\left(t\right)$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}\frac{tcos\left(t\right)}{1+{sin}^2\left(t\right)}=\left[tarctan\left(sin\left(t\right)\right)\right]-{\int }_0^{\frac{\pi }{2}}arctan\left(sin\left(t\right)\right)dt$$$$=\frac{{\pi }^2}{8}-{\int }_0^1\frac{arctan\left(r\right)}{\sqrt{1-r^2}}dr$$$$f\left(s\right)={\int }_0^1\frac{arctan\left(sr\right)}{\sqrt{1-r^2}}dr,f\left(0\right)=0$$$$f^{\prime }\left(s\right)={\int }_0^1\frac{r}{\sqrt{1-r^2}(1+s^2{r}^2)}dr,r=sin\left(x\right)$$$$f^{\prime }\left(s\right)={\int }_0^{\frac{\pi }{2}}\frac{sin\left(x\right)}{1+s^2{sin}^2\left(x\right)}={\int }_0^{\frac{\pi }{2}}\frac{sin\left(x\right)}{1+s^2-s^2{cos}^2\left(x\right)}$$$$=-\frac{1}{s\sqrt{1+s^2}}{\int }_0\frac{s(-sin\left(x\right))dx}{\sqrt{1+s^2}(1-{\left(\frac{s}{\sqrt{1+s^2}}cos\left(x\right)\right)}^2}$$$$=-\frac{1}{s\sqrt{1+s^2}}{\left[arcth\left(\frac{s}{\sqrt{1+s^2}}cos\left(x\right)\right)\right]}_0^{\frac{\pi }{2}}$$$$=\frac{1}{s\sqrt{1+s^2}}arcth\left(\frac{s}{\sqrt{1+s^2}}\right)=f^{\prime }\left(s\right)$$$$f\left(s\right)={\int }_0^s\frac{1}{s\sqrt{1+s^2}}arcth\left(\frac{s}{\sqrt{1+s^2}}\right)$$$$wewantf\left(1\right)$$$$f\left(1\right)={\int }_0^1\frac{1}{s\sqrt{1+s^2}}arcth\left(\frac{s}{\sqrt{1+s^2}}\right)$$$$lets=sh\left(t\right)\Rightarrow f\left(1\right)={\int }_0^{{sh}^{-}\left(1\right)}\frac{arcth\left(\frac{sh\left(t\right)}{ch\left(t\right)}\right)ch\left(t\right)dt}{sh\left(t\right)ch\left(t\right)}$$$$={\int }_0^{{sh}^{-}\left(1\right)}\frac{t}{sh\left(t\right)}dt...bypart$$$$\int \frac{1}{sh\left(t\right)}=ln(1-e^{-t})-ln(1+e^{-t})$$$$f\left(1\right)={\left[tln\left(\frac{1-e^{-t}}{1+e^{-t}}\right)\right]}_0^{{sh}^{-}\left(t\right)}-{\int }_0^{{sh}^{-}\left(1\right)}ln(1-e^{-t})dt$$$$+{\int }_0^{{sh}^{-}\left(t\right)}ln(1+e^{-t})$$$${\int }_0^{{sh}^{-}\left(1\right)}ln(1-e^{-t})dt,e^{-t}=w$$$$={\int }_1^{e^{-{sh}^{-}\left(1\right)}}\frac{ln(1-w)}{-w}dw={\int }_0^1\frac{ln(1-w)}{w}-{\int }_0^{e^{-{sh}^{-}\left(1\right)}}\frac{ln(1-w)}{w}dw$$$$={Li}_2\left(e^{-{sh}^{-}\left(1\right)}\right)-\mathit{L}{i}_2\left(1\right)$$$${\int }_0^{e^{-{sh}^{-}\left(1\right)}}ln(1+e^{-t})dt,e^{-t}=-w$$$$\Rightarrow {\int }_{-1}^{-e^{-{sh}^{-}\left(1\right)}}\frac{ln(1-w)dw}{-w}={Li}_2(-e^{-{sh}^{-}\left(1\right)})-{li}_2(-1)$$$$weget,a={sh}^{-}\left(1\right)$$$$f\left(1\right)=aln\left(\frac{1-e^{-a}}{1+e^{-a}}\right)+{Li}_2(-e^{-a})-{li}_2(-1)-{Li}_2\left(e^{-a}\right)+{Li}_2\left(1\right)$$$${Li}_2\left(1\right)=\zeta \left(1\right)=\frac{{\pi }^2}{6},{li}_2(-1)=-\frac{{\pi }^2}{12}$$$$f\left(1\right)=\frac{{\pi }^2}{4}+{Li}_2(-e^{-a})-{Li}_2\left(e^{-a}\right)+aln\left(\frac{1-e^{-a}}{1+e^{-a}}\right)$$$$\mathrm{\Omega }=\frac{{\pi }^2}{8}-f\left(1\right)..iwilltryifthereispossibility$$$$togive{Li}_2(...)-{Li}_2(...)byelementryfunction$$$$$$

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