Use-gamma-function-to-prove-i-0-8-cos-3-4x-dx-1-6-ii-0-6-cos-4-3-sin-2-6-d-5-192-

Question Number 90574 by niroj last updated on 24/Apr/20

$$\:\boldsymbol{\mathrm{Use}}\:\boldsymbol{\mathrm{gamma}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{prove}} \\ $$$$\:\:\left(\mathrm{i}\right)\:.\:\:\int_{\mathrm{0}} ^{\:\:\frac{\boldsymbol{\pi}}{\mathrm{8}}} \:\boldsymbol{\mathrm{cos}}^{\mathrm{3}} \mathrm{4}\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{dx}}=\:\frac{\mathrm{1}}{\mathrm{6}}. \\ $$$$\:\:\left(\boldsymbol{\mathrm{ii}}\right).\:\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{6}}} \:\boldsymbol{\mathrm{cos}}^{\mathrm{4}} \mathrm{3}\boldsymbol{\theta}\:\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \mathrm{6}\boldsymbol{\theta}\:\boldsymbol{\mathrm{d}\theta}\:=\:\frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{192}}. \\ $$

Commented by mathmax by abdo last updated on 24/Apr/20

$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{cos}^{\mathrm{3}} \left(\mathrm{4}{x}\right){dx}\:\Rightarrow\:{I}\:=_{\mathrm{4}{x}={t}} \:\:\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{3}} \left({t}\right)\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}} {t}\:\:{sint}\:{dt} \\ $$$${B}\left({p},{q}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{p}−\mathrm{1}} {x}\:{sin}^{\mathrm{2}{q}−\mathrm{1}} {xdx}\: \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} {t}\:{sin}^{\mathrm{2}} {t}\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{1}} {t}\:{sin}^{\mathrm{2}\left(\mathrm{1}\right)−\mathrm{1}} {tdt} \\ $$$$={B}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right).\Gamma\left(\mathrm{1}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)}\:{we}\:{have}\:\:\:\:\Gamma\left({n}\right)=\left({n}−\mathrm{1}\right)!\:\Rightarrow\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)\:\Rightarrow \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)\:=\frac{\mathrm{3}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:\Rightarrow{B}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\frac{\mathrm{3}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{2}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$

Commented by niroj last updated on 25/Apr/20

$$\mathrm{thank}\:\mathrm{you}\:. \\ $$

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