Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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.

$${you}\:{are}\:{welcome}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com