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Question Number 87409 by redmiiuser last updated on 04/Apr/20
∫(sinx)^(1/5) dx
$$\int\left({sinx}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} {dx} \\ $$
Commented by Prithwish Sen 1 last updated on 04/Apr/20
∫((sin^(1/5) x cosx)/(cosx)) dx put sinx = u^5 cosxdx = 5u^4 du =∫((5u^5 du)/( (√(1−u^(10) )))) =5∫u^5 (1−u^(10) )^(−(1/2)) dx ∣x∣≤1 now apply the formula of the series...
$$\int\frac{\mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{5}}} \mathrm{x}\:\mathrm{cosx}}{\mathrm{cosx}}\:\mathrm{dx}\:\:\:\mathrm{put}\:\mathrm{sinx}\:=\:\mathrm{u}^{\mathrm{5}} \:\:\mathrm{cosxdx}\:=\:\mathrm{5u}^{\mathrm{4}} \mathrm{du} \\ $$$$=\int\frac{\mathrm{5u}^{\mathrm{5}} \mathrm{du}}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{10}} \:}}\:\:=\mathrm{5}\int\mathrm{u}^{\mathrm{5}} \left(\mathrm{1}−\mathrm{u}^{\mathrm{10}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dx}\:\:\:\mid\mathrm{x}\mid\leqslant\mathrm{1} \\ $$$$\mathrm{now}\:\mathrm{apply}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}… \\ $$
Commented by redmiiuser last updated on 04/Apr/20
Yes mister you are correct.God bless you!
$${Yes}\:{mister}\:{you}\:{are} \\ $$$${correct}.{God}\:{bless}\:{you}! \\ $$
Answered by mind is power last updated on 04/Apr/20
=∫(t^(1/5) /( (√(1−t^2 ))))dx,t=sin(x) (1/( (√(1−t^2 ))))=Σ_(n≥0) (((2n)!)/(2^(2n) .(n!)^2 ))t^(2n) =∫Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 ))t^(2n+(1/5)) dt=Σ_(n≥0) (((2n)!5t^(2n+(6/5)) )/(2^(2n) (n!)^2 (10n+6))) =(1/2)t^(6/5) Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 (n+(3/5))))=(t^(6/5) /2)((5/3)+Σ_(n≥1) (((2n)!x^(2n) )/(2^(2n) (n!)^2 (n+(3/5))))) =((5t^(6/5) )/6)(1+Σ_(n≥1) ((2^n n!.Π_(k=0) ^(n−1) (2k+1).(3/5))/(2^(2n) (n!)^2 .(n+(3/5)))).t^(2n) ) =((5t^(6/5) )/6)(1+Σ_(n≥1) ((Π_(k=0) ^(n−1) ((1/2)+k).Π_(k=0) ^(n−1) ((3/5)+k))/(Π_(k=0) ^(n−1) ((8/5)+k))).(((t^2 )^n )/(n!))) =(5/6)t^(6/5) _2 F_1 ((1/2),(3/5);(8/5);t^2 )+c =(5/6)sin(x)^(6/5) _2 F_1 ((1/2),(3/5);(8/5);sin^2 (x))+c
$$=\int\frac{{t}^{\frac{\mathrm{1}}{\mathrm{5}}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dx},{t}={sin}\left({x}\right) \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} .\left({n}!\right)^{\mathrm{2}} }{t}^{\mathrm{2}{n}} \\ $$$$=\int\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{t}^{\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{5}}} {dt}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!\mathrm{5}{t}^{\mathrm{2}{n}+\frac{\mathrm{6}}{\mathrm{5}}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left(\mathrm{10}{n}+\mathrm{6}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\frac{\mathrm{6}}{\mathrm{5}}} \underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left({n}+\frac{\mathrm{3}}{\mathrm{5}}\right)}=\frac{{t}^{\frac{\mathrm{6}}{\mathrm{5}}} }{\mathrm{2}}\left(\frac{\mathrm{5}}{\mathrm{3}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\mathrm{2}{n}\right)!{x}^{\mathrm{2}{n}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} \left({n}+\frac{\mathrm{3}}{\mathrm{5}}\right)}\right) \\ $$$$=\frac{\mathrm{5}{t}^{\frac{\mathrm{6}}{\mathrm{5}}} }{\mathrm{6}}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}^{{n}} {n}!.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\mathrm{2}{k}+\mathrm{1}\right).\frac{\mathrm{3}}{\mathrm{5}}}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} .\left({n}+\frac{\mathrm{3}}{\mathrm{5}}\right)}.{t}^{\mathrm{2}{n}} \right) \\ $$$$=\frac{\mathrm{5}{t}^{\frac{\mathrm{6}}{\mathrm{5}}} }{\mathrm{6}}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{3}}{\mathrm{5}}+{k}\right)}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\frac{\mathrm{8}}{\mathrm{5}}+{k}\right)}.\frac{\left(\mathrm{t}^{\mathrm{2}} \right)^{{n}} }{{n}!}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{6}}{t}^{\frac{\mathrm{6}}{\mathrm{5}}} \:\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{5}};\frac{\mathrm{8}}{\mathrm{5}};\mathrm{t}^{\mathrm{2}} \right)+\mathrm{c} \\ $$$$=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{sin}\left(\mathrm{x}\right)^{\frac{\mathrm{6}}{\mathrm{5}}} \:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{5}};\frac{\mathrm{8}}{\mathrm{5}};{sin}^{\mathrm{2}} \left({x}\right)\right)+{c} \\ $$$$ \\ $$$$ \\ $$

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