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cos-2t-t-3-dt-




Question Number 50466 by afachri last updated on 16/Dec/18
   ∫ ((cos (2t))/t^3 ) dt = ....
$$\: \\ $$$$\int\:\frac{\mathrm{cos}\:\left(\mathrm{2}{t}\right)}{{t}^{\mathrm{3}} }\:{dt}\:=\:…. \\ $$
Commented by afachri last updated on 16/Dec/18
Would anybody give me a hand  on this ? i′m stuck.
$$\boldsymbol{\mathrm{Would}}\:\boldsymbol{\mathrm{anybody}}\:\boldsymbol{\mathrm{give}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{hand}} \\ $$$$\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{this}}\:?\:\boldsymbol{\mathrm{i}}'\boldsymbol{\mathrm{m}}\:\boldsymbol{\mathrm{stuck}}.\: \\ $$
Commented by maxmathsup by imad last updated on 16/Dec/18
at form of serie  we have cos(x)=Σ_(n=0) ^∞  (((−1)^n x^(2n) )/((2n)!)) ⇒  cos(2t) =Σ_(n=0) ^∞   (((−1)^n )/((2n)!)) 2^(2n)   t^(2n)  ⇒∫  ((cos(2t))/t^3 )dt  =Σ_(n=0) ^∞   (((−1)^(n ) 4^n )/((2n)!))∫ t^(2n−3) dt  =Σ_(n=0) ^∞  (−1)^n  (4^n /((2n)!(2n−2))) t^(2n−2)  +c .
$${at}\:{form}\:{of}\:{serie}\:\:{we}\:{have}\:{cos}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\:\Rightarrow \\ $$$${cos}\left(\mathrm{2}{t}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:\mathrm{2}^{\mathrm{2}{n}} \:\:{t}^{\mathrm{2}{n}} \:\Rightarrow\int\:\:\frac{{cos}\left(\mathrm{2}{t}\right)}{{t}^{\mathrm{3}} }{dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}\:} \mathrm{4}^{{n}} }{\left(\mathrm{2}{n}\right)!}\int\:{t}^{\mathrm{2}{n}−\mathrm{3}} {dt}\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{\mathrm{4}^{{n}} }{\left(\mathrm{2}{n}\right)!\left(\mathrm{2}{n}−\mathrm{2}\right)}\:{t}^{\mathrm{2}{n}−\mathrm{2}} \:+{c}\:. \\ $$$$ \\ $$
Commented by afachri last updated on 17/Dec/18
aha, it′s McLaurin series Sir. Thank u  very much Sir
$$\mathrm{aha},\:\mathrm{it}'\mathrm{s}\:\mathrm{McLaurin}\:\mathrm{series}\:\mathrm{Sir}.\:\mathrm{Thank}\:\mathrm{u} \\ $$$$\mathrm{very}\:\mathrm{much}\:\mathrm{Sir} \\ $$
Commented by maxmathsup by imad last updated on 17/Dec/18
you are welcome.
$${you}\:{are}\:{welcome}. \\ $$

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