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Question Number 170549 by mathlove last updated on 26/May/22

$$\int_{\mathrm{0}} ^{\mathrm{1}} {xarctan}^{\mathrm{6}} {x}\:{dx}=? \\$$

Answered by Mathspace last updated on 26/May/22

$${by}\:\rho{arts}\:\:{I}=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}^{\mathrm{6}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} − \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\frac{\mathrm{6}{arctanx}^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{4}} }{\mathrm{6}}\right)−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \left({arctanx}\right)^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$ $${changement}\:{arctanx}=\theta\:{give} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \frac{\left({arctanx}\right)^{\mathrm{5}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tan}^{\mathrm{2}} \theta×\frac{\theta^{\mathrm{5}} }{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{5}} {tan}^{\mathrm{2}} {x}\:{dx} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {x}−\mathrm{1}\right){x}^{\mathrm{5}} {dx} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{5}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{5}} {dx} \\$$ $$=−\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{6}} +\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{5}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx} \\$$ $${and}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{5}} \left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx} \\$$ $$=\left[{x}^{\mathrm{5}} {tanx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{5}{x}^{\mathrm{4}} {tanx}\:{dx} \\$$ $$=\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{5}} −\mathrm{5}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{4}} {tanx}\:{dx} \\$$ $${tanx}=\Sigma\:{a}_{\mathrm{2}{n}+\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \\$$ $$={a}_{\mathrm{1}} {x}\:+{a}_{\mathrm{3}} {x}^{\mathrm{3}} +... \\$$ $${f}\left({x}\right)={tanx}\:=\sum_{{i}} \:\frac{{f}^{\left({i}\right)} \left(\mathrm{0}\right)}{{i}!}{x}^{{i}} \\$$ $${f}\left({o}\right)=\mathrm{0}\:\:{f}^{'} \left({x}\right)=\mathrm{1}+{tan}^{\mathrm{2}} {x}\:\Rightarrow \\$$ $${f}^{'} \left({o}\right)=\mathrm{1} \\$$ $${f}^{\left(\mathrm{2}\right)} \left({x}\right)=\mathrm{2}{tanx}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\:\Rightarrow \\$$ $${f}^{\left(\mathrm{2}\right)} \left(\mathrm{0}\right)=\mathrm{0} \\$$ $${f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)=\mathrm{2}\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)+ \\$$ $$\mathrm{2}{tanx}\left(....\right)\:\Rightarrow{f}^{\left(\mathrm{3}\right)} \left(\mathrm{0}\right)=\mathrm{2}\:\Rightarrow \\$$ $${tanx}={x}\:+\frac{\mathrm{2}}{\mathrm{3}!}{x}^{\mathrm{3}} +... \\$$ $$={x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+...\:\Rightarrow \\$$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{4}} {tanx}\:{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({x}^{\mathrm{5}} +\frac{{x}^{\mathrm{4}} }{\mathrm{3}}+...\right){dx} \\$$ $$=\left[\frac{{x}^{\mathrm{6}} }{\mathrm{6}}+\frac{{x}^{\mathrm{5}} }{\mathrm{15}}+....\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{6}} +\frac{\mathrm{1}}{\mathrm{15}}\left(\frac{\pi}{\mathrm{4}}\right)^{\mathrm{5}} +.... \\$$

Commented byTawa11 last updated on 08/Oct/22

$$\mathrm{Great}\:\mathrm{sir} \\$$

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