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

 Use gamma function to prove    (i) .  āˆ«_0 ^(  (š›‘/8))  cos^3 4x dx= (1/6).    (ii). āˆ«_0 ^( (š›‘/6))  cos^4 3š›‰ sin^2 6š›‰ dš›‰ = ((5š›‘)/(192)).

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

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

I =āˆ«_0 ^(Ļ€/8)  cos^3 (4x)dx ā‡’ I =_(4x=t)    (1/4)āˆ«_0 ^(Ļ€/2)  cos^3 (t) dt  =(1/4) āˆ«_0 ^(Ļ€/2)   cos^2 t  sint dt  B(p,q) =āˆ«_0 ^(Ļ€/2)  cos^(2pāˆ’1) x sin^(2qāˆ’1) xdx   ā‡’āˆ«_0 ^(Ļ€/2)  cos^2 t sin^2 t dt =āˆ«_0 ^(Ļ€/2)  cos^(2((3/2))āˆ’1) t sin^(2(1)āˆ’1) tdt  =B((3/2),1) =((Ī“((3/2)).Ī“(1))/(Ī“((3/2)+1))) we have    Ī“(n)=(nāˆ’1)! ā‡’Ī“(1)=1  Ī“(x+1)=xĪ“(x) ā‡’  Ī“((3/2)+1) =(3/2)Ī“((3/2)) ā‡’B((3/2),1) =((Ī“((3/2)))/((3/2)Ī“((3/2)))) =(2/3) ā‡’  I =(1/4)Ɨ(2/3) =(1/6)

$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \:{cos}^{\mathrm{3}} \left(\mathrm{4}{x}\right){dx}\:\Rightarrow\:{I}\:=_{\mathrm{4}{x}={t}} \:\:\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{3}} \left({t}\right)\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:{cos}^{\mathrm{2}} {t}\:\:{sint}\:{dt} \\ $$$${B}\left({p},{q}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{p}āˆ’\mathrm{1}} {x}\:{sin}^{\mathrm{2}{q}āˆ’\mathrm{1}} {xdx}\: \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}} {t}\:{sin}^{\mathrm{2}} {t}\:{dt}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)āˆ’\mathrm{1}} {t}\:{sin}^{\mathrm{2}\left(\mathrm{1}\right)āˆ’\mathrm{1}} {tdt} \\ $$$$={B}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right).\Gamma\left(\mathrm{1}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)}\:{we}\:{have}\:\:\:\:\Gamma\left({n}\right)=\left({n}āˆ’\mathrm{1}\right)!\:\Rightarrow\Gamma\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right)\:\Rightarrow \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}\right)\:=\frac{\mathrm{3}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:\Rightarrow{B}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{1}\right)\:=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\frac{\mathrm{3}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{4}}Ɨ\frac{\mathrm{2}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$

Commented by niroj last updated on 25/Apr/20

thank you .

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

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