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Question Number 118668 by bemath last updated on 19/Oct/20

$$\int \frac{x\mid \sin x\mid }{1+\cos {}^{2}x}dx?$$

Answered by Lordose last updated on 19/Oct/20

$$\mathrm{\Omega }=\int \frac{\mathrm{x}\mid \mathrm{sinx}\mid }{{\sin }^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{\Omega }=\int \mathrm{x}\mid \mathrm{cosecx}\mid \mathrm{dx}$$$$\mathrm{IBP}\{\mathrm{u}=\mathrm{x}\mathrm{dv}=\mathrm{cosecxdx}\}$$$$\mathrm{\Omega }=\mathrm{xln}\mid \tan \left(\frac{\mathrm{x}}{2}\right)\mid -\int \ln \mid \tan \left(\frac{\mathrm{x}}{2}\right)\mid \mathrm{dx}$$$$\mathrm{\Delta }=\int \ln \mid \tan \left(\frac{\mathrm{x}}{2}\right)\mid \mathrm{dx}$$$$\mathrm{\Delta }=2\int \ln \mid \tan \left(\mathrm{u}\right)\mid \mathrm{du}\{\mathrm{u}=\frac{\mathrm{x}}{2}\}$$$$\mathrm{\Delta }=2\int \ln \mid \frac{-{\mathrm{ie}}^{\mathrm{iu}}(1-{\mathrm{e}}^{-2\mathrm{iu}})}{{\mathrm{e}}^{\mathrm{iu}}(1+{\mathrm{e}}^{-2\mathrm{iu}})}\mid \mathrm{du}$$$$\mathrm{\Delta }=2(\int \ln \mid -\mathbf{i}\mid \mathrm{du}+\int \ln \mid 1-{\mathrm{e}}^{2\mathrm{iu}}\mid \mathrm{du}-\int \ln \mid 1+{\mathrm{e}}^{-2\mathrm{iu}}\mid \mathrm{du})$$$$\mathrm{y}={\mathrm{e}}^{-2\mathrm{iu}}\Rightarrow \mathrm{du}=\frac{\mathrm{idy}}{2\mathrm{y}}$$$$\mathrm{\Delta }=2(\frac{\mathrm{i}\pi \mathrm{u}}{2}+\frac{\mathrm{i}}{2}\int \frac{\ln \mid 1-\mathrm{y}\mid }{\mathrm{y}}\mathrm{dy}-\frac{\mathrm{i}}{2}\int \frac{\ln \mid 1-\mathrm{y}\mid }{\mathrm{y}}\mathrm{dy}$$$$\mathrm{\Delta }=\mathrm{i}\pi \mathrm{u}+{\mathrm{iLi}}_2\left(\mathrm{y}\right)-{\mathrm{iLi}}_2(-\mathrm{y})+\mathrm{C}$$$$\mathrm{\Delta }=\mathrm{i}(\pi \mathrm{u}+{\mathrm{Li}}_2\left({\mathrm{e}}^{-2\mathrm{iu}}\right)-{\mathrm{Li}}_2(-{\mathrm{e}}^{-2\mathrm{iu}}))+\mathrm{C}$$$$\mathrm{\Omega }=\mathrm{xln}\mid \tan \left(\frac{\mathrm{x}}{2}\right)\mid -\mathrm{i}(\frac{\pi \mathrm{x}}{2}+{\mathrm{Li}}_2\left({\mathrm{e}}^{-\mathrm{ix}}\right)-{\mathrm{Li}}_2(-{\mathrm{e}}^{\mathrm{ix}}))+\mathrm{C}$$$$\mathrm{Check}\mathrm{for}\mathrm{errors}!$$

Commented by bemath last updated on 19/Oct/20

$$thankyousir$$

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