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Question Number 45498 by Sanjarbek last updated on 13/Oct/18

Commented by Meritguide1234 last updated on 13/Oct/18

$$notsolvable$$

Commented by MJS last updated on 13/Oct/18

$$\mathrm{possible}\mathrm{but}\mathrm{not}\mathrm{elementary}.{\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{post}\mathrm{it}\mathrm{later}$$

Commented by maxmathsup by imad last updated on 13/Oct/18

$$letf\left(t\right)=\int ln\left(sin\left(tx\right)\right)dx\Rightarrow f^{{}^{\prime }}\left(t\right)=\int \frac{xcos\left(tx\right)}{sin\left(tx\right)}dx$$$${=}_{tx=u}\int \frac{u}{t}\frac{cos\left(u\right)}{sin\left(u\right)}\frac{du}{t}=\frac{1}{t^2}\int u\frac{cos\left(u\right)}{sin\left(u\right)}dualsochangementtan\left(\frac{u}{2}\right)=\alpha$$$$givef\left(t\right)=\frac{1}{t^2}\int 2arctan\left(\alpha \right)\frac{\frac{1-{\alpha }^2}{1+{\alpha }^2}}{\frac{2\alpha }{1+{\alpha }^2}}\frac{2d\alpha }{1+{\alpha }^2}$$$$=\frac{4}{t^2}\int \frac{arctan\left(\alpha \right)}{1+{\alpha }^2}d\alpha letthenintroducetheparametricfunction$$$$\phi \left(u\right)=\int \frac{arctan\left(u\alpha \right)}{1+{\alpha }^2}d\alpha \Rightarrow {\phi }^{{}^{\prime }}\left(u\right)=\int \frac{\alpha }{(1+{\alpha }^2)(1+u^2{\alpha }^2)}d\alpha letdecompose$$$$F\left(\alpha \right)=\frac{a\alpha +b}{1+{\alpha }^2}+\frac{c\alpha +d}{1+u^2{\alpha }^2}....becontinued...$$$$$$$$$$

Answered by MJS last updated on 14/Oct/18

$$\int \ln \sin xdx=\int \ln (-\frac{\mathrm{i}}{2}({\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}))dx=$$$$[t=\mathrm{i}x\to dx=-\mathrm{i}dt$$$$=-\mathrm{i}\int \ln (-\frac{\mathrm{i}}{2}({\mathrm{e}}^t-{\mathrm{e}}^{-t}))dt=$$$$=-\mathrm{i}\int \ln (-\frac{\mathrm{i}}{2})dt-\mathrm{i}\int \ln ({\mathrm{e}}^t-{\mathrm{e}}^{-t})dt$$$$\mathrm{the}\mathrm{first}\mathrm{one}\mathrm{is}\mathrm{easy}:$$$$-\mathrm{i}\int \ln (-\frac{\mathrm{i}}{2})dt=-\mathrm{i}\ln (-\frac{\mathrm{i}}{2})\int dt=-\mathrm{i}\ln (-\frac{\mathrm{i}}{2})t=$$$$=\ln (-\frac{\mathrm{i}}{2})x$$$$\mathrm{the}\mathrm{second}\mathrm{one}:$$$$-\mathrm{i}\int \ln ({\mathrm{e}}^t-{\mathrm{e}}^{-t})dt=$$$$\left[\begin{array}{c}\int f^{\prime }g=fg-\int {fg}^{\prime }\\ f^{\prime }=1\to f=t\\ g=\ln ({\mathrm{e}}^t-{\mathrm{e}}^{-t})\to g^{\prime }=\frac{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}{{\mathrm{e}}^t-{\mathrm{e}}^{-t}}\end{array}\right]$$$$=-\mathrm{i}t\ln ({\mathrm{e}}^t-{\mathrm{e}}^{-t})+\mathrm{i}\int t\frac{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}{{\mathrm{e}}^t-{\mathrm{e}}^{-t}}dt$$$$\mathrm{the}\mathrm{first}\mathrm{term}:$$$$-\mathrm{i}t\ln ({\mathrm{e}}^t-{\mathrm{e}}^{-t})=x\ln ({\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x})$$$$\mathrm{the}\mathrm{second}\mathrm{term}:$$$$\mathrm{i}\int t\frac{{\mathrm{e}}^t+{\mathrm{e}}^{-t}}{{\mathrm{e}}^t-{\mathrm{e}}^{-t}}dt=\mathrm{i}\int t\frac{{\mathrm{e}}^{2t}+1}{{\mathrm{e}}^{2t}-1}dt=$$$$[u={\mathrm{e}}^{2t}-1\to dt=\frac{{\mathrm{e}}^{-2t}}{2}du]$$$$=\frac{\mathrm{i}}{4}\int \frac{(u+2)\ln (u+1)}{u(u+1)}du=$$$$=\frac{\mathrm{i}}{2}\int \frac{\ln (u+1)}{u}du+\frac{\mathrm{i}}{4}\int \frac{\ln (u+1)}{u+1}du$$$$\mathrm{the}\mathrm{first}\mathrm{one}:$$$$\frac{\mathrm{i}}{2}\int \frac{\ln (u+1)}{u}du=$$$$[v=-u\to du=-dv]$$$$=\frac{\mathrm{i}}{2}\int \frac{\ln (1-v)}{v}dv=-\frac{\mathrm{i}}{2}\int -\frac{\ln (1-v)}{v}dv=$$$$\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{special}\mathrm{integral}\left(\mathrm{dilogarithm}\right)$$$$=-\frac{\mathrm{i}}{2}{\mathrm{Li}}_{2}v=-\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(-u)=-\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(1-{\mathrm{e}}^{2t})=$$$$=-\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(1-{\mathrm{e}}^{2\mathrm{i}x})$$$$\mathrm{the}\mathrm{second}\mathrm{one}:$$$$\frac{\mathrm{i}}{4}\int \frac{\ln (u+1)}{u+1}du=$$$$[w=\ln (u+1)\to du=(u+1)dw]$$$$=\frac{\mathrm{i}}{4}\int wdw=\frac{\mathrm{i}}{8}{w}^2=\frac{\mathrm{i}}{8}{\ln }^2(u+1)=\frac{\mathrm{i}}{8}{\ln }^2\left({\mathrm{e}}^{2t}\right)=$$$$=\frac{\mathrm{i}}{2}{t}^2=-\frac{\mathrm{i}}{2}{x}^2$$$$\mathrm{so}\mathrm{we}\mathrm{have}$$$$\int \ln \sin xdx=$$$$=x\ln (-\frac{\mathrm{i}}{2})+x\ln ({\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x})-\frac{\mathrm{i}}{2}{\mathrm{Li}}_2(1-{\mathrm{e}}^{2\mathrm{i}x})-\frac{\mathrm{i}}{2}{x}^2+C$$$$\mathrm{please}\mathrm{check}...$$

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