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Question Number 139371 by mohammad17 last updated on 26/Apr/21

by use Gamma function prove (1) ∫_0 ^( (π/8)) cos^3 4xdx=(1/6) (2) ∫_0 ^( π) sin^6 ((x/2))cos^8 ((x/2))dx=((5π)/2^(11) )

$${by}\:{use}\:{Gamma}\:{function}\:{prove}\: \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{8}}} {cos}^{\mathrm{3}} \mathrm{4}{xdx}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\:\pi} {sin}^{\mathrm{6}} \left(\frac{{x}}{\mathrm{2}}\right){cos}^{\mathrm{8}} \left(\frac{{x}}{\mathrm{2}}\right){dx}=\frac{\mathrm{5}\pi}{\mathrm{2}^{\mathrm{11}} } \\ $$

Answered by Dwaipayan Shikari last updated on 26/Apr/21

(x/2)=u ⇒2∫_0 ^(π/2) sin^6 (u)cos^8 (u)du=((Γ((7/2))Γ((9/2)))/(Γ(8)))=(((5/2).(3/2).(3/2).(5/2).(7/2).(1/2).(1/2)π)/(7!)) =(π/2^7 ).((5.5.3.3)/(6!))=(π/2^8 ).((3.5)/(4!))=((5π)/2^(11) )

$$\frac{{x}}{\mathrm{2}}={u} \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{6}} \left({u}\right){cos}^{\mathrm{8}} \left({u}\right){du}=\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{8}\right)}=\frac{\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{5}}{\mathrm{2}}.\frac{\mathrm{7}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}\pi}{\mathrm{7}!} \\ $$$$=\frac{\pi}{\mathrm{2}^{\mathrm{7}} }.\frac{\mathrm{5}.\mathrm{5}.\mathrm{3}.\mathrm{3}}{\mathrm{6}!}=\frac{\pi}{\mathrm{2}^{\mathrm{8}} }.\frac{\mathrm{3}.\mathrm{5}}{\mathrm{4}!}=\frac{\mathrm{5}\pi}{\mathrm{2}^{\mathrm{11}} } \\ $$

Commented by mohammad17 last updated on 26/Apr/21

thank you sir and number (1) can you solve

$${thank}\:{you}\:{sir}\:{and}\:{number}\:\left(\mathrm{1}\right)\:{can}\:{you}\:{solve} \\ $$

Answered by Dwaipayan Shikari last updated on 26/Apr/21

4x=u ⇒(1/4)∫_0 ^(π/2) cos^3 (u)du=(1/4).((Γ(2)Γ((1/2)))/(2Γ((5/2))))=(1/4).((√π)/((3(√π))/4)).(1/2)=(1/6)

$$\mathrm{4}{x}={u} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{3}} \left({u}\right){du}=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{4}}.\frac{\sqrt{\pi}}{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}}.\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by mohammad17 last updated on 26/Apr/21

sir can you clear this steb how get Γ(2) and Γ((1/2)) from cos^3 (u)

$${sir}\:{can}\:{you}\:{clear}\:{this}\:{steb}\:{how}\:{get} \\ $$$$ \\ $$$$\Gamma\left(\mathrm{2}\right)\:{and}\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:{from}\:{cos}^{\mathrm{3}} \left({u}\right) \\ $$

Commented by mohammad17 last updated on 26/Apr/21

thank you sir

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

Commented by Dwaipayan Shikari last updated on 26/Apr/21

∫_0 ^(π/2) cos^(2n−1) x sin^(2α−1) x dx=((Γ(n)Γ(α))/(2Γ(n+α))) ∫_0 ^(π/2) cos^3 x dx=∫_0 ^(π/2) cos^(2.2−1) x sin^(2((1/2))−1) xdx=((Γ(2)Γ((1/2)))/(2Γ(2+(1/2))))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{n}−\mathrm{1}} {x}\:{sin}^{\mathrm{2}\alpha−\mathrm{1}} {x}\:{dx}=\frac{\Gamma\left({n}\right)\Gamma\left(\alpha\right)}{\mathrm{2}\Gamma\left({n}+\alpha\right)} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{3}} {x}\:{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}.\mathrm{2}−\mathrm{1}} {x}\:{sin}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} {xdx}=\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$

Answered by Ar Brandon last updated on 26/Apr/21

I=∫_0 ^(π/8) cos^3 4xdx , u=4x =(1/4)∫_0 ^(π/2) cos^3 udu=(1/8)β(2,(1/2)) =(1/8)∙((Γ(2)Γ((1/2)))/(Γ((5/2))))=(1/8)∙((√π)/((3/2)∙(1/2)∙(√π)))=(1/6) J=∫_0 ^π sin^6 ((x/2))cos^8 ((x/2))dx , x=2v =2∫_0 ^(π/2) sin^6 vcos^8 vdv=β((7/2),(9/2)) =((Γ((7/2))Γ((9/2)))/(Γ(8)))=(1/(7!))∙((5/2)∙(3/2)∙((√π)/2))^2 ∙(7/2) =((225π)/(720×2^7 ))=((5π)/(2048))=((5π)/2^(11) )

$$\mathcal{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{8}}} \mathrm{cos}^{\mathrm{3}} \mathrm{4xdx}\:,\:\mathrm{u}=\mathrm{4x} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{3}} \mathrm{udu}=\frac{\mathrm{1}}{\mathrm{8}}\beta\left(\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\Gamma\left(\mathrm{2}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\sqrt{\pi}}{\frac{\mathrm{3}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\sqrt{\pi}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$ \\ $$$$\mathcal{J}=\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\mathrm{6}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}^{\mathrm{8}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{dx}\:,\:\mathrm{x}=\mathrm{2v} \\ $$$$\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{6}} \mathrm{vcos}^{\mathrm{8}} \mathrm{vdv}=\beta\left(\frac{\mathrm{7}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{8}\right)}=\frac{\mathrm{1}}{\mathrm{7}!}\centerdot\left(\frac{\mathrm{5}}{\mathrm{2}}\centerdot\frac{\mathrm{3}}{\mathrm{2}}\centerdot\frac{\sqrt{\pi}}{\mathrm{2}}\right)^{\mathrm{2}} \centerdot\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{225}\pi}{\mathrm{720}×\mathrm{2}^{\mathrm{7}} }=\frac{\mathrm{5}\pi}{\mathrm{2048}}=\frac{\mathrm{5}\pi}{\mathrm{2}^{\mathrm{11}} } \\ $$

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