Question Number 81217 by M±th+et£s last updated on 10/Feb/20 | ||
Answered by mind is power last updated on 10/Feb/20 | ||
$${i}\:{will}\:{poste}\:{all}\:{solution}\:{later}! \\ $$$${for}\:\mathrm{2}{nd}\:{Somme}\:{mistacks} \\ $$$${i}\:{found} \\ $$$${f}\left({x}\right)=\int\frac{\sqrt{{tan}\left(\mathrm{2}{x}\right)}}{{sin}\left({x}\right)}{dx}=\mathrm{2}\sqrt{\mathrm{2}.{tan}\left({x}\right)}.\:\:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{9}}{\mathrm{8}};{tan}^{\mathrm{4}} \left({x}\right)\right)+{c} \\ $$$${tan}\left(\mathrm{2}{x}\right)=\frac{\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)},\:\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\left[\right.\right. \\ $$$${sin}\left({x}\right)=\sqrt{{sin}^{\mathrm{2}} \left({x}\right)},=\sqrt{\frac{{tan}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}} \\ $$$$\int\frac{\sqrt{{tan}\left(\mathrm{2}{x}\right)}}{{sin}\left({x}\right)}{dx}=\int\sqrt{\frac{\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)}}.\sqrt{\frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{{tan}^{\mathrm{2}} \left({x}\right)}}{dx} \\ $$$$=\int\sqrt{\frac{\mathrm{2}}{{tan}\left({x}\right)}}.\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right).\frac{{dx}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}!\right)}{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} }{x}^{\mathrm{2}{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{tg}^{\mathrm{4}{n}} \left({x}\right) \\ $$$$\int\sqrt{\frac{\mathrm{2}}{{tan}\left({x}\right)}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right).\frac{{dx}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}} \\ $$$$=\int\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }.{tg}^{\mathrm{4}{n}−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right)\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)\right)\sqrt{\mathrm{2}}{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\sqrt{\mathrm{2}}\int{tg}^{\mathrm{4}{n}−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right)\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)\right){dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\sqrt{\mathrm{2}}.\frac{{tg}^{\mathrm{4}{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{4}{n}+\frac{\mathrm{1}}{\mathrm{2}}}+{c}=\mathrm{2}\sqrt{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{tg}^{\mathrm{4}{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{8}{n}+\mathrm{1}\right)}+{c} \\ $$$$ \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}.\sqrt{{tg}\left({x}\right)}.\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!!.{tg}^{\mathrm{4}{n}} \left({x}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{8}{n}+\mathrm{1}\right)}+{c} \\ $$$$\left(\mathrm{2}{n}\right)!=\left(\mathrm{2}{n}−\mathrm{1}\right)!!.\mathrm{2}^{{n}} .{n}! \\ $$$${f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!.\mathrm{2}^{{n}} {n}!}{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} \left(\mathrm{8}{n}+\mathrm{1}\right)}.\left({tg}^{\mathrm{4}} \left({x}\right)\right)^{{n}} \right)+{c} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} }.\frac{\mathrm{1}}{\mathrm{8}{n}+\mathrm{1}}.\frac{\left({tg}^{\mathrm{4}} \left({x}\right)\right)^{{n}} }{{n}!}\right)+{c} \\ $$$$\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} }=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)......\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}−\mathrm{1}\right) \\ $$$$\mathrm{8}{n}+\mathrm{1}=\mathrm{8}\left({n}+\frac{\mathrm{9}}{\mathrm{8}}−\mathrm{1}\right)\Leftrightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{8}{n}+\mathrm{1}}=\frac{\frac{\mathrm{1}}{\mathrm{8}}.......\left(\frac{\mathrm{1}}{\mathrm{8}}+{n}−\mathrm{1}\right)}{\frac{\mathrm{9}}{\mathrm{8}}........\left(\frac{\mathrm{9}}{\mathrm{8}}+{n}−\mathrm{1}\right)} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{8}}+{k}−\mathrm{1}\right)}{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\mathrm{9}}{\mathrm{8}}+{k}−\mathrm{1}\right)}.\frac{\left({tg}^{\mathrm{4}} \left({x}\right)\right)^{{n}} }{{n}!}\right)+{c} \\ $$$$\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({a}_{{n}} \right)\left({b}_{{n}} \right)}{{c}_{{n}} }.\frac{{x}^{{n}} }{{n}!}=\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b},{c};{x}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{9}}{\mathrm{8}};{tg}^{\mathrm{4}} \left({x}\right)\right)+{c}\:\:,{c}\:{canstante} \\ $$$$\int\frac{\sqrt{{tan}\left(\mathrm{2}{x}\right)}}{{sin}\left({x}\right)}{dx}=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{9}}{\mathrm{8}};{tg}^{\mathrm{4}} \left({x}\right)\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by M±th+et£s last updated on 10/Feb/20 | ||
$${great}\:{sir}\:{god}\:{bless}\:{you}\: \\ $$ | ||
Commented by mind is power last updated on 10/Feb/20 | ||
$${thank}\:{you}\:{sir}\:{for}\:{posting} \\ $$$${somm}\:{hypergeometric}\:{Function} \\ $$ | ||
Answered by M±th+et£s last updated on 11/Feb/20 | ||
Commented by mind is power last updated on 12/Feb/20 | ||
$${nice} \\ $$ | ||