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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 2nd Somme mistacks i found f(x)=∫((√(tan(2x)))/(sin(x)))dx=2(√(2.tan(x))). _2 F_1 ((1/2),(1/8),(9/8);tan^4 (x))+c tan(2x)=((2tan(x))/(1−tan^2 (x))), ∀x∈[0,(π/4)[ sin(x)=(√(sin^2 (x))),=(√((tan^2 (x))/(1+tan^2 (x)))) ∫((√(tan(2x)))/(sin(x)))dx=∫(√((2tan(x))/(1−tan^2 (x)))).(√((1+tan^2 (x))/(tan^2 (x))))dx =∫(√(2/(tan(x)))).(1+tan^2 (x)).(dx/(√(1−tan^4 (x)))) (1/(√(1−x^2 )))=Σ_(n≥0) (((2n!))/(2^(2n) .(n!)^2 ))x^(2n) ⇒(1/(√(1−tan^4 (x))))=Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 ))tg^(4n) (x) ∫(√(2/(tan(x))))(1+tan^2 (x)).(dx/(√(1−tan^4 (x)))) =∫Σ_(n=0) ^(+∞) (((2n)!)/(2^(2n) (n!)^2 )).tg^(4n−(1/2)) (x)(1+tg^2 (x))(√2)dx =Σ_(n=0) ^(+∞) (((2n)!)/(2^(2n) (n!)^2 ))(√2)∫tg^(4n−(1/2)) (x)(1+tg^2 (x))dx =Σ_(n=0) ^(+∞) (((2n)!)/(2^(2n) (n!)^2 ))(√2).((tg^(4n+(1/2)) )/(4n+(1/2)))+c=2(√2)Σ_(n≥0) (((2n)!tg^(4n+(1/2)) )/(2^(2n) (n!)^2 (8n+1)))+c =2(√2).(√(tg(x))).Σ_(n≥0) (((2n)!!.tg^(4n) (x))/(2^(2n) (n!)^2 (8n+1)))+c (2n)!=(2n−1)!!.2^n .n! f(x)=2(√(2tan(x))) (1+Σ_(n≥1) (((2n−1)!!.2^n n!)/(2^(2n) .(n!)^2 (8n+1))).(tg^4 (x))^n )+c =2(√(2tan(x))) (1+Σ_(n≥1) (((2n−1)!!)/2^n ).(1/(8n+1)).(((tg^4 (x))^n )/(n!)))+c (((2n−1)!!)/2^n )=((1/2))......((1/2)+n−1) 8n+1=8(n+(9/8)−1)⇔ (1/(8n+1))=(((1/8).......((1/8)+n−1))/((9/8)........((9/8)+n−1))) ⇒f(x)=2(√(2tan(x)))(1+Σ_(n≥1) ((Π_(k=1) ^n ((1/2)+k−1)((1/8)+k−1))/(Π_(k=1) ^n ((9/8)+k−1))).(((tg^4 (x))^n )/(n!)))+c 1+Σ_(n≥1) (((a_n )(b_n ))/c_n ).(x^n /(n!))= _2 F_1 (a,b,c;x) f(x)=2(√(2tan(x))) _2 F_1 ((1/2),(1/8),(9/8);tg^4 (x))+c ,c canstante ∫((√(tan(2x)))/(sin(x)))dx=2(√(2tan(x))) 2F1((1/2),(1/8),(9/8);tg^4 (x))+c

$${i}\:{will}\:{poste}\:{all}\:{solution}\:{later}! \\ $$$${for}\:\mathrm{2}{nd}\:{Somme}\:{mistacks} \\ $$$${i}\:{found} \\ $$$${f}\left({x}\right)=\int\frac{\sqrt{{tan}\left(\mathrm{2}{x}\right)}}{{sin}\left({x}\right)}{dx}=\mathrm{2}\sqrt{\mathrm{2}.{tan}\left({x}\right)}.\:\:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{9}}{\mathrm{8}};{tan}^{\mathrm{4}} \left({x}\right)\right)+{c} \\ $$$${tan}\left(\mathrm{2}{x}\right)=\frac{\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)},\:\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\left[\right.\right. \\ $$$${sin}\left({x}\right)=\sqrt{{sin}^{\mathrm{2}} \left({x}\right)},=\sqrt{\frac{{tan}^{\mathrm{2}} \left({x}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}} \\ $$$$\int\frac{\sqrt{{tan}\left(\mathrm{2}{x}\right)}}{{sin}\left({x}\right)}{dx}=\int\sqrt{\frac{\mathrm{2}{tan}\left({x}\right)}{\mathrm{1}−{tan}^{\mathrm{2}} \left({x}\right)}}.\sqrt{\frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{{tan}^{\mathrm{2}} \left({x}\right)}}{dx} \\ $$$$=\int\sqrt{\frac{\mathrm{2}}{{tan}\left({x}\right)}}.\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right).\frac{{dx}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}!\right)}{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} }{x}^{\mathrm{2}{n}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{tg}^{\mathrm{4}{n}} \left({x}\right) \\ $$$$\int\sqrt{\frac{\mathrm{2}}{{tan}\left({x}\right)}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)\right).\frac{{dx}}{\sqrt{\mathrm{1}−{tan}^{\mathrm{4}} \left({x}\right)}} \\ $$$$=\int\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }.{tg}^{\mathrm{4}{n}−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right)\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)\right)\sqrt{\mathrm{2}}{dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\sqrt{\mathrm{2}}\int{tg}^{\mathrm{4}{n}−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right)\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)\right){dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\sqrt{\mathrm{2}}.\frac{{tg}^{\mathrm{4}{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{4}{n}+\frac{\mathrm{1}}{\mathrm{2}}}+{c}=\mathrm{2}\sqrt{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{tg}^{\mathrm{4}{n}+\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{8}{n}+\mathrm{1}\right)}+{c} \\ $$$$ \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}.\sqrt{{tg}\left({x}\right)}.\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!!.{tg}^{\mathrm{4}{n}} \left({x}\right)}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{8}{n}+\mathrm{1}\right)}+{c} \\ $$$$\left(\mathrm{2}{n}\right)!=\left(\mathrm{2}{n}−\mathrm{1}\right)!!.\mathrm{2}^{{n}} .{n}! \\ $$$${f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!.\mathrm{2}^{{n}} {n}!}{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} \left(\mathrm{8}{n}+\mathrm{1}\right)}.\left({tg}^{\mathrm{4}} \left({x}\right)\right)^{{n}} \right)+{c} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} }.\frac{\mathrm{1}}{\mathrm{8}{n}+\mathrm{1}}.\frac{\left({tg}^{\mathrm{4}} \left({x}\right)\right)^{{n}} }{{n}!}\right)+{c} \\ $$$$\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}} }=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)......\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}−\mathrm{1}\right) \\ $$$$\mathrm{8}{n}+\mathrm{1}=\mathrm{8}\left({n}+\frac{\mathrm{9}}{\mathrm{8}}−\mathrm{1}\right)\Leftrightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{8}{n}+\mathrm{1}}=\frac{\frac{\mathrm{1}}{\mathrm{8}}.......\left(\frac{\mathrm{1}}{\mathrm{8}}+{n}−\mathrm{1}\right)}{\frac{\mathrm{9}}{\mathrm{8}}........\left(\frac{\mathrm{9}}{\mathrm{8}}+{n}−\mathrm{1}\right)} \\ $$$$\Rightarrow{f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{8}}+{k}−\mathrm{1}\right)}{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\mathrm{9}}{\mathrm{8}}+{k}−\mathrm{1}\right)}.\frac{\left({tg}^{\mathrm{4}} \left({x}\right)\right)^{{n}} }{{n}!}\right)+{c} \\ $$$$\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left({a}_{{n}} \right)\left({b}_{{n}} \right)}{{c}_{{n}} }.\frac{{x}^{{n}} }{{n}!}=\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b},{c};{x}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{9}}{\mathrm{8}};{tg}^{\mathrm{4}} \left({x}\right)\right)+{c}\:\:,{c}\:{canstante} \\ $$$$\int\frac{\sqrt{{tan}\left(\mathrm{2}{x}\right)}}{{sin}\left({x}\right)}{dx}=\mathrm{2}\sqrt{\mathrm{2}{tan}\left({x}\right)}\:\mathrm{2}{F}\mathrm{1}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{9}}{\mathrm{8}};{tg}^{\mathrm{4}} \left({x}\right)\right)+{c} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 10/Feb/20

great sir god bless you

$${great}\:{sir}\:{god}\:{bless}\:{you}\: \\ $$

Commented by mind is power last updated on 10/Feb/20

thank you sir for posting somm hypergeometric Function

$${thank}\:{you}\:{sir}\:{for}\:{posting} \\ $$$${somm}\:{hypergeometric}\:{Function} \\ $$

Answered by M±th+et£s last updated on 11/Feb/20

Commented by mind is power last updated on 12/Feb/20

nice

$${nice} \\ $$

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