Question Number 201980 by Calculusboy last updated on 17/Dec/23
Answered by Sutrisno last updated on 18/Dec/23
$${misal}\:{x}^{\mathrm{2}} ={t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{dx}=\frac{{dt}}{\mathrm{2}{x}} \\ $$$$=\int{x}.{x}^{\mathrm{2}} {cot}\left({x}^{\mathrm{2}} \right)\frac{{dt}}{\mathrm{2}{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{tcot}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({t}.{ln}\mid{sint}\mid−\int{ln}\mid{sint}\mid{dt}\right)\:\:\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$${I}=\int{ln}\mid{sint}\mid{dt} \\ $$$${I}=\int{ln}\mid\mathrm{2}.{sin}\frac{\mathrm{1}}{\mathrm{2}}{t}.{cos}\frac{\mathrm{1}}{\mathrm{2}}{t}\mid{dt} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:+\int\mid{sin}\frac{\mathrm{1}}{\mathrm{2}}{t}.{cos}\frac{\mathrm{1}}{\mathrm{2}}{t}\mid{dt} \\ $$$${misal}\:{t}=\pi−{y}\:\rightarrow\:−{dt}={dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:+\int\mid{sin}\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−{y}\right).{cos}\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−{y}\right)\mid.−{dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:−\int\mid{cos}\frac{\mathrm{1}}{\mathrm{2}}\left({y}\right).{sin}\frac{\mathrm{1}}{\mathrm{2}}\left({y}\right)\mid{dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\mid{sin}\left({y}\right)\mid{dy} \\ $$$${I}=\int{ln}\mathrm{2}\:{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}{I} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{I}={tln}\mathrm{2}\:\rightarrow\:{I}=\frac{\mathrm{2}}{\mathrm{3}}{tln}\mathrm{2} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({t}.{ln}\mid{sint}\mid−\frac{\mathrm{2}}{\mathrm{3}}{t}.{ln}\mathrm{2}\right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} .{ln}\mid{sinx}^{\mathrm{2}} \mid−\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{2}} .{ln}\mathrm{2}\right)+{c} \\ $$$$ \\ $$$$ \\ $$
Commented by Calculusboy last updated on 18/Dec/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$