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Question Number 97361 by student work last updated on 07/Jun/20

$$\int \frac{\mathrm{xdx}}{\sin {}^2\mathrm{x}-3}=?$$

Commented by student work last updated on 07/Jun/20

$$\mathrm{who}\mathrm{is}\mathrm{intellagent}?$$

Commented by mr W last updated on 07/Jun/20

$$i{don}^{\prime }tknowwhoisintelligent.$$$$butaccordingtothisquestionofyou,$$$$iknowatleastwhoisnotintelligent.$$

Commented by som(math1967) last updated on 07/Jun/20

����

Commented by student work last updated on 07/Jun/20

$$\mathrm{solve}$$

Commented by student work last updated on 07/Jun/20

$$\mathrm{who}\mathrm{can}?$$

Commented by Tinku Tara last updated on 07/Jun/20

$$\mathrm{Hi}\mathrm{student}\mathrm{work}$$$$\mathrm{please}\mathrm{stop}\mathrm{putting}\mathrm{comments}$$$$\mathrm{such}\mathrm{as}\mathrm{who}\mathrm{is}\mathrm{intelligent},\mathrm{solve}$$$$\mathrm{who}\mathrm{can}\mathrm{etc}.$$$$\mathrm{Polite}\mathrm{comments}\mathrm{like}\mathrm{pls}\mathrm{help},$$$$\mathrm{i}\mathrm{need}\mathrm{help}\mathrm{for}\mathrm{assignment}$$$$\mathrm{are}\mathrm{fine}.$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{I}=\int \frac{\mathrm{xdx}}{{\sin }^2\mathrm{x}-3}=\int \frac{\mathrm{xdx}}{-2-{\cos }^2\mathrm{x}}=-\int \frac{\mathrm{xdx}}{2+\frac{1}{1+{\tan }^2\mathrm{x}}}$$$$=-\int \frac{\mathrm{x}(1+{\tan }^2\mathrm{x})}{3+{2\tan }^2\mathrm{x}}\mathrm{dx}\mathrm{changement}\mathrm{tanx}=\mathrm{u}\mathrm{give}$$$$\mathrm{I}=-\int \frac{(1+{\mathrm{u}}^2)\mathrm{arctanu}}{3+{2\mathrm{u}}^2}\frac{\mathrm{du}}{1+{\mathrm{u}}^2}=-\int \frac{\mathrm{arctanu}}{3+{2\mathrm{u}}^2}\mathrm{du}$$$$=-\frac{1}{3}\int \frac{\mathrm{arctanu}}{1+\frac{2}{3}{\mathrm{u}}^2}\mathrm{du}\mathrm{and}\mathrm{changement}\sqrt{\frac{2}{3}}\mathrm{u}=\mathrm{z}\mathrm{give}$$$$\mathrm{I}=-\frac{1}{3}\int \frac{\arctan \left(\sqrt{\frac{3}{2}}\mathrm{z}\right)}{1+{\mathrm{z}}^2}\times \sqrt{\frac{3}{2}}\mathrm{dz}=-\frac{1}{\sqrt{6}}\int \frac{\arctan \left(\sqrt{\frac{3}{2}}\mathrm{z}\right)}{1+{\mathrm{z}}^2}\mathrm{dz}$$$$\mathrm{let}\mathrm{put}\mathrm{f}\left(\alpha \right)={\int }_0^{\mathrm{x}}\frac{\arctan \left(\alpha \mathrm{z}\right)}{1+{\mathrm{z}}^2}\mathrm{dz}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\alpha \right)={\int }_0^{\mathrm{x}}\frac{\mathrm{z}}{(1+{\mathrm{z}}^2)(1+{\alpha }^2{\mathrm{z}}^2)}\mathrm{dz}{=}_{\alpha \mathrm{z}=\mathrm{u}}{\int }_0^{\alpha \mathrm{x}}\frac{\mathrm{u}}{\alpha (1+\frac{{\mathrm{u}}^2}{{\alpha }^2})(1+{\mathrm{u}}^2)}\frac{\mathrm{du}}{\alpha }$$$$={\int }_0^{\alpha \mathrm{x}}\frac{\mathrm{udu}}{({\mathrm{u}}^2+{\alpha }^2)({\mathrm{u}}^2+1)}\mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{solvable}\mathrm{by}\mathrm{decomposition}$$$$...\mathrm{be}\mathrm{continued}....$$

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