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Question Number 98338 by  M±th+et+s last updated on 13/Jun/20

∫cos(x^(18) ) dx

$$\int{cos}\left({x}^{\mathrm{18}} \right)\:{dx} \\ $$$$ \\ $$

Answered by smridha last updated on 13/Jun/20

let x=k^(1/(18)) so =(1/(18))∫k^((1/(18))−1) coskdk =(1/(18))Σ_(n=0) ^∞ (((−1)^n )/((2n)!))∫k^(2n+(1/(18))−1) dk =(1/(18))Σ_(n=0) ^∞ (((−1)^n )/((2n)!))(x^(36n+1) /((2n+(1/(18)))))+c

$$\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{k}}^{\frac{\mathrm{1}}{\mathrm{18}}} \boldsymbol{{so}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\int\boldsymbol{{k}}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} \boldsymbol{{coskdk}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}\boldsymbol{{n}}\right)!}\int\boldsymbol{{k}}^{\mathrm{2}\boldsymbol{{n}}+\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} \boldsymbol{{dk}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\left(\mathrm{2}\boldsymbol{{n}}\right)!}\frac{\boldsymbol{{x}}^{\mathrm{36}\boldsymbol{{n}}+\mathrm{1}} }{\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{18}}\right)}+\boldsymbol{{c}} \\ $$

Commented by  M±th+et+s last updated on 13/Jun/20

Commented by  M±th+et+s last updated on 13/Jun/20

and i have another solution with using incomplete gamma function i will post it later

$${and}\:{i}\:{have}\:{another}\:{solution}\:{with}\:{using} \\ $$$${incomplete}\:{gamma}\:{function}\:{i}\:{will} \\ $$$${post}\:{it}\:{later} \\ $$

Commented by smridha last updated on 13/Jun/20

well thanks a lot..

$$\boldsymbol{{well}}\:\boldsymbol{{thanks}}\:\boldsymbol{{a}}\:\boldsymbol{{lot}}.. \\ $$

Commented by aurpeyz last updated on 13/Jun/20

how did you get the summation stuff

$$\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{summation}\:\mathrm{stuff} \\ $$

Commented by smridha last updated on 13/Jun/20

series expanssion of cosx.

$$\boldsymbol{{series}}\:\boldsymbol{{expanssion}}\:\boldsymbol{{of}}\:\boldsymbol{{cosx}}. \\ $$

Answered by  M±th+et+s last updated on 13/Jun/20

cos(x^(18) )=((e^(ix^(18) ) +e^(−ix^(18) ) )/2) I=(1/2)∫e^(ix^(18) ) dx+ (1/2)∫e^(−ix^(18) ) dx I_1 =∫e^(−ix^(18) ) dx ; let ix^(18) =y −iy=x^(18) dx=(((−i^(1/(18)) ))/(18))y^((1/(18))−1) dy =(((−i^(1/(18)) ))/(18))∫e^(−y) y^((1/(18))−1) dy =−(−i)^(1/(18)) (((−1)/(18)))∫e^(−y) y^((1/(18))−1) dy but ((−1)/(18))∫_0 ^y e^(−y) y^((1/(18))−1) dy+c= =(1/(18))(∫_0 ^∞ e^(−y) y^((1/(18))−1) dy−∫_y ^∞ e^(−y) y^((1/(18))−1) dy)+c −(1/(18))(∫_0 ^∞ e^(−y) y^((1/(18))−1) dy−∫_(0 ) ^∞ e^(−y) y^((1/(18))−1) )+c =(1/(18))Γ((1/(18)),y) I_1 =((−(−i)^(1/(18)) )/(18))Γ((1/(18)),y) I_1 =((−x.Γ((1/(18)),ix^(18) ))/(18((ix^(18) ))^(1/(18)) ))+c_1 I_2 =∫e^(ix^(18) ) dx;let ix^(18) =−y dx=(((i)^(1/(18)) )/(18))y^((1/(18))−1) dy I_2 =(((i)^(1/(18)) )/(18))∫e^(−y) y^((1/(18))−1) dy =(((i)^(1/(18)) )/(18))Γ((1/(18)),−ix^(18) )+c_2 =((−x.Γ((1/(18)),−ix^(18) ))/(18((−ix^(18) ))^(1/(18)) ))+c where Γ(b,x) is the incomplete gamma function I=((−x[((ix^(18) ))^(1/(18)) .Γ((1/(18)),−ix^(18) )+((−ix^(18) ))^(1/(18)) .Γ((1/(18)),ix^(18) ))/(36 (x^(36) )^(1/(18)) ))+c

$${cos}\left({x}^{\mathrm{18}} \right)=\frac{{e}^{{ix}^{\mathrm{18}} } +{e}^{−{ix}^{\mathrm{18}} } }{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{{ix}^{\mathrm{18}} } \:{dx}+\:\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{−{ix}^{\mathrm{18}} } {dx} \\ $$$${I}_{\mathrm{1}} =\int{e}^{−{ix}^{\mathrm{18}} } {dx}\:;\:{let}\:\:{ix}^{\mathrm{18}} ={y}\:\:−{iy}={x}^{\mathrm{18}} \\ $$$$\:{dx}=\frac{\left(−{i}^{\frac{\mathrm{1}}{\mathrm{18}}} \right)}{\mathrm{18}}{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$$=\frac{\left(−{i}^{\frac{\mathrm{1}}{\mathrm{18}}} \right)}{\mathrm{18}}\int{e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$$=−\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} \left(\frac{−\mathrm{1}}{\mathrm{18}}\right)\int{e}^{−{y}} \:{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$${but}\:\frac{−\mathrm{1}}{\mathrm{18}}\int_{\mathrm{0}} ^{{y}} {e}^{−{y}} \:{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}+{c}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}−\int_{{y}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}\right)+{c} \\ $$$$−\frac{\mathrm{1}}{\mathrm{18}}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}−\int_{\mathrm{0}\:} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} \right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{y}\right) \\ $$$${I}_{\mathrm{1}} =\frac{−\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{y}\right) \\ $$$${I}_{\mathrm{1}} =\frac{−{x}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{ix}^{\mathrm{18}} \right)}{\mathrm{18}\sqrt[{\mathrm{18}}]{{ix}^{\mathrm{18}} }}+{c}_{\mathrm{1}} \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\int{e}^{{ix}^{\mathrm{18}} } {dx};{let}\:\:{ix}^{\mathrm{18}} =−{y}\:{dx}=\frac{\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$${I}_{\mathrm{2}} =\frac{\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}\int{e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$$=\frac{\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},−{ix}^{\mathrm{18}} \right)+{c}_{\mathrm{2}} \\ $$$$=\frac{−{x}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},−{ix}^{\mathrm{18}} \right)}{\mathrm{18}\sqrt[{\mathrm{18}}]{−{ix}^{\mathrm{18}} }}+{c} \\ $$$$ \\ $$$${where}\:\Gamma\left({b},{x}\right)\:{is}\:{the}\:{incomplete}\:{gamma} \\ $$$${function} \\ $$$$ \\ $$$${I}=\frac{−{x}\left[\sqrt[{\mathrm{18}}]{{ix}^{\mathrm{18}} }.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},−{ix}^{\mathrm{18}} \right)+\sqrt[{\mathrm{18}}]{−{ix}^{\mathrm{18}} }.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{ix}^{\mathrm{18}} \right)\right.}{\mathrm{36}\:\sqrt[{\mathrm{18}}]{{x}^{\mathrm{36}} }}+{c} \\ $$$$ \\ $$$$ \\ $$

Commented by smridha last updated on 13/Jun/20

very nice solution..

$$\boldsymbol{{very}}\:\boldsymbol{{nice}}\:\boldsymbol{{solution}}.. \\ $$

Commented by  M±th+et+s last updated on 13/Jun/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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