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cosh-n-x-dx-please-help-




Question Number 5882 by sanusihammed last updated on 03/Jun/16
∫cosh^n (x) dx    please help.
coshn(x)dxpleasehelp.
Answered by Yozzii last updated on 03/Jun/16
cosh^n (x)=(((e^x +e^(−x) )/2))^n    n∈N+{0} (assumed)  cosh^n (x)=(1/2^n )Σ_(k=0) ^n { ((n),(k) )(e^x )^(n−k) (e^(−x) )^k }  (Binomial theorem)  cosh^n x=2^(−n) Σ_(k=0) ^n { ((n),(k) )e^(x(n−2k)) }  For term independent of x, ⇒n−2k=0  or k=(n/2) where k∈Z. So,if n is even  then a term independent of x exists  and is given by 2^(−n)  ((n),((n/2)) ). Define  the set R={n,k∈Z^≥ , n≡0(mod 2)∣0≤k≤n and k≠(n/2)}.  ⇒∫cosh^n xdx=∫{2^(−n) Σ_(k∈R)  ((n),(k) )e^(x(n−2k)) +2^(−n)  ((n),((n/2)) )}dx  ∫cosh^n xdx=2^(−n) {Σ_(k∈R) [ ((n),(k) )(e^(x(n−2k)) /(n−2k))]+x ((n),((n/2)) )}+C  This is valid for non−zero even n.  If n=0⇒∫cosh^n xdx=x+C  −−−−−−−−−−−−−−−−−−−−−−−−−−  If n is odd then k=(n/2)∉Z, so that   no term independent of x exists.  ∴ ∫cosh^n xdx=2^(−n) Σ_(k=0) ^n [ ((n),(k) )(e^(x(n−2k)) /(n−2k))]+C  −−−−−−−−−−−−−−−−−−−−−−−−−−  In all, for n∈N+{0}, where R={k,n∈Z^≥ , n≡0(mod 2)∣0≤k≤n and k≠(n/2)}  ∫cosh^n (x)dx= { ((2^(−n) {Σ_(k∈R) [ ((n),(k) )(e^(x(n−2k)) /(n−2k))]+ ((n),((n/2)) )x}+C      if n≡0(mod 2) and n≥2)),((2^(−n) Σ_(k=0) ^n [ ((n),(k) )(e^(x(n−2k)) /(n−2k))]+C                                if n is odd)),((x+C                                                                      if n=0)) :}  C is a constant of integration.
coshn(x)=(ex+ex2)nnN+{0}(assumed)coshn(x)=12nnk=0{(nk)(ex)nk(ex)k}(Binomialtheorem)coshnx=2nnk=0{(nk)ex(n2k)}Fortermindependentofx,n2k=0ork=n2wherekZ.So,ifniseventhenatermindependentofxexistsandisgivenby2n(nn/2).DefinethesetR={n,kZ,n0(mod2)0knandkn2}.coshnxdx={2nkR(nk)ex(n2k)+2n(nn/2)}dxcoshnxdx=2n{kR[(nk)ex(n2k)n2k]+x(nn/2)}+CThisisvalidfornonzeroevenn.Ifn=0coshnxdx=x+CIfnisoddthenk=n2Z,sothatnotermindependentofxexists.coshnxdx=2nnk=0[(nk)ex(n2k)n2k]+CInall,fornN+{0},whereR={k,nZ,n0(mod2)0knandkn2}coshn(x)dx={2n{kR[(nk)ex(n2k)n2k]+(nn/2)x}+Cifn0(mod2)andn22nnk=0[(nk)ex(n2k)n2k]+Cifnisoddx+Cifn=0Cisaconstantofintegration.
Commented by Yozzii last updated on 03/Jun/16
Suppose n∈Z^− ⇒let n=−r where r∈Z^+ .  ⇒I(n)=I(−r)=∫cosh^(−r) xdx  I(−r)=∫sech^r x dx=2^r ∫(dx/((e^x +e^(−x) )^r ))  I(−r)=2^r ∫(e^(2rx) /((1+e^(2x) )^r ))dx=2^r ∫((e^x /(1+e^(2x) )))^r dx  u=e^x ⇒du=e^x dx⇒dx=(1/u)du  ⇒I(−r)=2^r ∫((u/(1+u^2 )))^r (1/u)du  I(−r)=2^r ∫(u^(r−1) /((1+u^2 )^r ))du  ...
SupposenZletn=rwhererZ+.I(n)=I(r)=coshrxdxI(r)=sechrxdx=2rdx(ex+ex)rI(r)=2re2rx(1+e2x)rdx=2r(ex1+e2x)rdxu=exdu=exdxdx=1uduI(r)=2r(u1+u2)r1uduI(r)=2rur1(1+u2)rdu
Commented by sanusihammed last updated on 03/Jun/16
Thanks sir.
Thankssir.
Commented by sanusihammed last updated on 03/Jun/16
Commented by sanusihammed last updated on 03/Jun/16
Please check that result. how come ?
Pleasecheckthatresult.howcome?
Commented by Yozzii last updated on 03/Jun/16
The function F(...) is a hypergeometric  function. My level of knowledge and  umderstanding of Mathematics is  insufficient to prove that result.   Sorry...
ThefunctionF()isahypergeometricfunction.MylevelofknowledgeandumderstandingofMathematicsisinsufficienttoprovethatresult.Sorry
Commented by sanusihammed last updated on 03/Jun/16
  Thanks i really appreiate
Thanksireallyappreiate

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