Question Number 98338 by M±th+et+s last updated on 13/Jun/20
$$\int{cos}\left({x}^{\mathrm{18}} \right)\:{dx} \\ $$$$ \\ $$
Answered by smridha last updated on 13/Jun/20
$$\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{k}}^{\frac{\mathrm{1}}{\mathrm{18}}} \boldsymbol{{so}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\int\boldsymbol{{k}}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} \boldsymbol{{coskdk}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}\boldsymbol{{n}}\right)!}\int\boldsymbol{{k}}^{\mathrm{2}\boldsymbol{{n}}+\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} \boldsymbol{{dk}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\left(\mathrm{2}\boldsymbol{{n}}\right)!}\frac{\boldsymbol{{x}}^{\mathrm{36}\boldsymbol{{n}}+\mathrm{1}} }{\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{18}}\right)}+\boldsymbol{{c}} \\ $$
Commented by M±th+et+s last updated on 13/Jun/20
Commented by M±th+et+s last updated on 13/Jun/20
$${and}\:{i}\:{have}\:{another}\:{solution}\:{with}\:{using} \\ $$$${incomplete}\:{gamma}\:{function}\:{i}\:{will} \\ $$$${post}\:{it}\:{later} \\ $$
Commented by smridha last updated on 13/Jun/20
$$\boldsymbol{{well}}\:\boldsymbol{{thanks}}\:\boldsymbol{{a}}\:\boldsymbol{{lot}}.. \\ $$
Commented by aurpeyz last updated on 13/Jun/20
$$\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{summation}\:\mathrm{stuff} \\ $$
Commented by smridha last updated on 13/Jun/20
$$\boldsymbol{{series}}\:\boldsymbol{{expanssion}}\:\boldsymbol{{of}}\:\boldsymbol{{cosx}}. \\ $$
Answered by M±th+et+s last updated on 13/Jun/20
$${cos}\left({x}^{\mathrm{18}} \right)=\frac{{e}^{{ix}^{\mathrm{18}} } +{e}^{−{ix}^{\mathrm{18}} } }{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{{ix}^{\mathrm{18}} } \:{dx}+\:\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{−{ix}^{\mathrm{18}} } {dx} \\ $$$${I}_{\mathrm{1}} =\int{e}^{−{ix}^{\mathrm{18}} } {dx}\:;\:{let}\:\:{ix}^{\mathrm{18}} ={y}\:\:−{iy}={x}^{\mathrm{18}} \\ $$$$\:{dx}=\frac{\left(−{i}^{\frac{\mathrm{1}}{\mathrm{18}}} \right)}{\mathrm{18}}{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$$=\frac{\left(−{i}^{\frac{\mathrm{1}}{\mathrm{18}}} \right)}{\mathrm{18}}\int{e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$$=−\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} \left(\frac{−\mathrm{1}}{\mathrm{18}}\right)\int{e}^{−{y}} \:{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$${but}\:\frac{−\mathrm{1}}{\mathrm{18}}\int_{\mathrm{0}} ^{{y}} {e}^{−{y}} \:{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}+{c}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}−\int_{{y}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}\right)+{c} \\ $$$$−\frac{\mathrm{1}}{\mathrm{18}}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy}−\int_{\mathrm{0}\:} ^{\infty} {e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} \right)+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{y}\right) \\ $$$${I}_{\mathrm{1}} =\frac{−\left(−{i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{y}\right) \\ $$$${I}_{\mathrm{1}} =\frac{−{x}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{ix}^{\mathrm{18}} \right)}{\mathrm{18}\sqrt[{\mathrm{18}}]{{ix}^{\mathrm{18}} }}+{c}_{\mathrm{1}} \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\int{e}^{{ix}^{\mathrm{18}} } {dx};{let}\:\:{ix}^{\mathrm{18}} =−{y}\:{dx}=\frac{\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}{y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$${I}_{\mathrm{2}} =\frac{\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}\int{e}^{−{y}} {y}^{\frac{\mathrm{1}}{\mathrm{18}}−\mathrm{1}} {dy} \\ $$$$=\frac{\left({i}\right)^{\frac{\mathrm{1}}{\mathrm{18}}} }{\mathrm{18}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},−{ix}^{\mathrm{18}} \right)+{c}_{\mathrm{2}} \\ $$$$=\frac{−{x}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},−{ix}^{\mathrm{18}} \right)}{\mathrm{18}\sqrt[{\mathrm{18}}]{−{ix}^{\mathrm{18}} }}+{c} \\ $$$$ \\ $$$${where}\:\Gamma\left({b},{x}\right)\:{is}\:{the}\:{incomplete}\:{gamma} \\ $$$${function} \\ $$$$ \\ $$$${I}=\frac{−{x}\left[\sqrt[{\mathrm{18}}]{{ix}^{\mathrm{18}} }.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},−{ix}^{\mathrm{18}} \right)+\sqrt[{\mathrm{18}}]{−{ix}^{\mathrm{18}} }.\Gamma\left(\frac{\mathrm{1}}{\mathrm{18}},{ix}^{\mathrm{18}} \right)\right.}{\mathrm{36}\:\sqrt[{\mathrm{18}}]{{x}^{\mathrm{36}} }}+{c} \\ $$$$ \\ $$$$ \\ $$
Commented by smridha last updated on 13/Jun/20
$$\boldsymbol{{very}}\:\boldsymbol{{nice}}\:\boldsymbol{{solution}}.. \\ $$
Commented by M±th+et+s last updated on 13/Jun/20
$${thank}\:{you}\:{sir} \\ $$