Integration Questions

Question Number 130598 by Lordose last updated on 27/Jan/21

$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx} \\$$

Answered by mindispower last updated on 27/Jan/21

$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}\right).{d}\left({x}^{\mathrm{3}} \right)}{\mathrm{1}+\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} } \\$$$$=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\$$$${x}\rightarrow\frac{\mathrm{1}}{{x}} \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\frac{\pi}{\mathrm{2}}−{tan}^{−} \left({x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{{tan}^{−} \left({x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\$$$${we}\:{get}\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\$$

Commented by mnjuly1970 last updated on 27/Jan/21

$${very}\:{nice}\:{solution} \\$$$$\:{sir}\:\:{power}... \\$$

Answered by mathmax by abdo last updated on 27/Jan/21

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} \:\mathrm{arctan}\left(\mathrm{x}\right)}{\mathrm{1}+\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{2}} }\left(\mathrm{dx}^{\mathrm{3}} \right) \\$$$$=\frac{\mathrm{1}}{\mathrm{3}}\left\{\left[\:\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)\mathrm{arctanx}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\right\} \\$$$$\Rightarrow\mathrm{3I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\$$$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=_{\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}} \:\:−\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{arctant}^{\mathrm{3}} }{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\left(−\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\$$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{t}^{\mathrm{3}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }−\Phi\:\Rightarrow \\$$$$\mathrm{2}\Phi\:=\frac{\pi}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow\Phi=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\mathrm{3I}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\mathrm{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{24}} \\$$