Menu Close

show-that-e-sin-x-dx-n-0-1-n-cos-x-sin-x-n-1-sin-x-2-n-2-1-2-2F-1-1-2-1-n-2-3-2-cos-x-2-c-notice-2F-1-is-special-function-called-hyperge

Tinku Tara 0 Comments
Pin




Question Number 87793 by M±th+et£s last updated on 06/Apr/20
show that ∫e^(sin(x)) dx= −Σ_(n=0) ^∞ (1/(n!))[ cos(x)∗(sin(x))^(n+1) ∗[(sin(x))^2 ]^((((−n)/2)−(1/2))) ∗ 2F_1 [(1/2),((1−n)/2);(3/2);(cos(x))^2 ] ]+c notice\2F_1 is special function called hypergeometric function
$${show}\:{that} \\ $$$$\int{e}^{{sin}\left({x}\right)} \:{dx}= \\ $$$$−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\left[\:{cos}\left({x}\right)\ast\left({sin}\left({x}\right)\right)^{{n}+\mathrm{1}} \ast\left[\left({sin}\left({x}\right)\right)^{\mathrm{2}} \right]^{\left(\frac{−{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)} \ast\:\mathrm{2}{F}_{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}−{n}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};\left({cos}\left({x}\right)\right)^{\mathrm{2}} \right]\:\right]+{c} \\ $$$$ \\ $$$${notice}\backslash\mathrm{2}{F}_{\mathrm{1}} \:{is}\:{special}\:{function}\:{called}\:{hypergeometric}\:{function} \\ $$
Answered by mind is power last updated on 06/Apr/20
∫sin^m (x)dx =∫(u^m /( (√(1−u^2 ))))du∫Σ_(n≥0) u^m .(((2n)!u^(2n) )/(2^(2n) (n!)^2 ))du=Σ_(n≥0) ∫(((2n)!)/(2^(2n) (n!)^2 ))u^(2n+m) du =Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 )).(u^(2n+m+1) /((2n+m+1)))+c =u^(m+1) ((1/(m+1))+Σ_(n≥1) ((2^(2n) Π_(k=0) ^(n−1) (k+(1/2)))/(2Π_(k=0) ^(n−1) (k+((m+3)/2))))(u^(2n) /(n!))) =u^(m+1) =(u^(m+1) /(m+1)).Σ_(n≥0) ((Π_(k=0) ^(n−1) (k+(1/2)).Π_(k=0) ^(n−1) (k+((m+1)/2)))/(Π_(k=0) ^(n−1) (k+((m+3)/2)))).(u^(2n) /(n!)) =(u^(m+1) /(m+1)).2F_1 ((1/2),((m+1)/2);((m+3)/2);u^2 ) e^(sin(u)) =Σ_(m≥0) ((sin^m (u))/(m!)) ∫e^(sin(u)) du=∫Σ_(m≥0) ((sin^m (u))/(m!))du =Σ_(m≥0) (1/(m!))∫sin^m (u)du=Σ_(m≥0) ((sin^(m+1) (u))/(m!(m+1))).2F_1 ((1/2),((m+1)/2);((m+3)/2);sin^2 (u)) =Σ_(m≥0) ((sin^(m+1) (u))/((m+1)!))2F_1 ((1/2),((m+1)/2);((m+3)/2);sin^2 (u))+c may bee mistaks
$$\int{sin}^{{m}} \left({x}\right){dx} \\ $$$$=\int\frac{{u}^{{m}} }{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du}\int\underset{{n}\geqslant\mathrm{0}} {\sum}{u}^{{m}} .\frac{\left(\mathrm{2}{n}\right)!{u}^{\mathrm{2}{n}} }{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{du}=\underset{{n}\geqslant\mathrm{0}} {\sum}\int\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }{u}^{\mathrm{2}{n}+{m}} {du} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }.\frac{{u}^{\mathrm{2}{n}+{m}+\mathrm{1}} }{\left(\mathrm{2}{n}+{m}+\mathrm{1}\right)}+{c} \\ $$$$={u}^{{m}+\mathrm{1}} \left(\frac{\mathrm{1}}{{m}+\mathrm{1}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}^{\mathrm{2}{n}} \underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{{m}+\mathrm{3}}{\mathrm{2}}\right)}\frac{{u}^{\mathrm{2}{n}} }{{n}!}\right) \\ $$$$={u}^{{m}+\mathrm{1}} \\ $$$$=\frac{{u}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}.\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{{m}+\mathrm{1}}{\mathrm{2}}\right)}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{{m}+\mathrm{3}}{\mathrm{2}}\right)}.\frac{{u}^{\mathrm{2}{n}} }{{n}!} \\ $$$$=\frac{{u}^{{m}+\mathrm{1}} }{{m}+\mathrm{1}}.\mathrm{2}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{m}+\mathrm{1}}{\mathrm{2}};\frac{{m}+\mathrm{3}}{\mathrm{2}};{u}^{\mathrm{2}} \right) \\ $$$${e}^{{sin}\left({u}\right)} =\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{sin}^{{m}} \left({u}\right)}{{m}!} \\ $$$$\int{e}^{{sin}\left({u}\right)} {du}=\int\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{sin}^{{m}} \left({u}\right)}{{m}!}{du} \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{m}!}\int{sin}^{{m}} \left({u}\right){du}=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{sin}^{{m}+\mathrm{1}} \left({u}\right)}{{m}!\left({m}+\mathrm{1}\right)}.\mathrm{2}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{m}+\mathrm{1}}{\mathrm{2}};\frac{{m}+\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \left({u}\right)\right) \\ $$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{{sin}^{{m}+\mathrm{1}} \left({u}\right)}{\left({m}+\mathrm{1}\right)!}\mathrm{2}{F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{{m}+\mathrm{1}}{\mathrm{2}};\frac{{m}+\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \left({u}\right)\right)+{c} \\ $$$${may}\:{bee}\:{mistaks}\: \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
may bee i make a mistake i will post my solution
$${may}\:{bee}\:{i}\:{make}\:{a}\:{mistake}\:{i}\:{will}\:{post}\:{my} \\ $$$${solution} \\ $$
Answered by M±th+et£s last updated on 06/Apr/20
e^u =Σ_(n=0) ^∞ , u=sin(x)⇒⇒e^(sin(x)) =Σ_(n=0) ^∞ (((sin(x))^n )/(n!)) I=∫e^(sin(x)) dx=∫Σ_(n=0) ^∞ (((sin(x))^n )/(n!))dx=Σ_(n=0) ^∞ ∫(sin(x))^n dx ∫(sin(x))^n dx=−cos(x)∗(sin(x))^(n+1) ∗[sin(x)^2 ]^(((−n)/2)−(1/2)) ∗2F_1 [(1/2),((1−n)/2);(3/2);(cos(x))^2 ]o ∫e^(sin(x)) dx=−Σ_(n=0) ^∞ (1/(n!))[cos(x)∗sin(x)^(n+1) ∗[(sin(x))^2 ]^(((−n)/2)−(1/2)) ∗2F_1 [(1/2),((1−n)/2);(3/2);(icos(x)^2 )]]
$${e}^{{u}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:,\:{u}={sin}\left({x}\right)\Rightarrow\Rightarrow{e}^{{sin}\left({x}\right)} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({sin}\left({x}\right)\right)^{{n}} }{{n}!} \\ $$$${I}=\int{e}^{{sin}\left({x}\right)} {dx}=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({sin}\left({x}\right)\right)^{{n}} }{{n}!}{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int\left({sin}\left({x}\right)\right)^{{n}} {dx} \\ $$$$\int\left({sin}\left({x}\right)\right)^{{n}} {dx}=−{cos}\left({x}\right)\ast\left({sin}\left({x}\right)\right)^{{n}+\mathrm{1}} \ast\left[{sin}\left({x}\right)^{\mathrm{2}} \right]^{\frac{−{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \ast\mathrm{2}{F}_{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}−{n}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};\left({cos}\left({x}\right)\right)^{\mathrm{2}} \right]{o} \\ $$$$\int{e}^{{sin}\left({x}\right)} {dx}=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!}\left[{cos}\left({x}\right)\ast{sin}\left({x}\right)^{{n}+\mathrm{1}} \ast\left[\left({sin}\left({x}\right)\right)^{\mathrm{2}} \right]^{\frac{−{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \ast\mathrm{2}{F}_{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}−{n}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};\left({icos}\left({x}\right)^{\mathrm{2}} \right)\right]\right] \\ $$
Commented by M±th+et£s last updated on 06/Apr/20
there is a typo its [(cos(x))^2 ] not [(icos(x)^2 )]
$${there}\:{is}\:{a}\:{typo} \\ $$$${its}\:\left[\left({cos}\left({x}\right)\right)^{\mathrm{2}} \right]\:{not}\:\left[\left({icos}\left({x}\right)^{\mathrm{2}} \right)\right] \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *