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Question Number 82560 by niroj last updated on 22/Feb/20
Use gamma function to prove (i) ∫_0 ^(π/4) sin^4 x 2x dx = ((3𝛑−4)/(192)). (ii) ∫_0 ^(π/6) cos^4 3𝛉 sin^2 6θ = ((5π)/(192)).
$$\:\boldsymbol{\mathrm{U}}\mathrm{se}\:\mathrm{gamma}\:\mathrm{function}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\:\:\left(\mathrm{i}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}\:\mathrm{2}\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{dx}}\:=\:\frac{\mathrm{3}\boldsymbol{\pi}−\mathrm{4}}{\mathrm{192}}. \\ $$$$\:\:\left(\mathrm{ii}\right)\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\boldsymbol{\mathrm{cos}}^{\mathrm{4}} \mathrm{3}\boldsymbol{\theta}\:\mathrm{sin}^{\mathrm{2}} \mathrm{6}\theta\:=\:\frac{\mathrm{5}\pi}{\mathrm{192}}. \\ $$
Answered by M±th+et£s last updated on 22/Feb/20
N.B:Γ(z+(1/2))=(((2z−1)!!)/2^z )(√π) β=((Γ(x)Γ(y))/(Γ(x+y)))=∫_0 ^(π/2) cos^(2x−1) θ sin^(2x−1) θ dθ I=∫_0 ^(π/6) cos^4 3θ sin^2 6θ dθ =^(θ→3θ) (1/3)∫_0 ^(π/2) cos^4 θ sin^2 2θ dθ =(4/3)∫_0 ^(π/2) cos^6 θ sin^2 θ dθ=(2/3)β((7/2),(3/2)) I=(2/3) ((Γ((7/2),(3/2)))/(Γ(5)))=(2/3) ((Γ(3+(1/2))Γ(1+(1/2)))/(Γ(5))) =(2/3) (((((5!!)/2^3 )(√π) )(((1!!)/2^1 ) (√π) ))/(4!))=((2π)/3) (((((15)/8))((1/2)))/(24)) I=((5π)/(192))
$${N}.{B}:\Gamma\left({z}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\left(\mathrm{2}{z}−\mathrm{1}\right)!!}{\mathrm{2}^{{z}} }\sqrt{\pi}\:\:\:\:\beta=\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{x}−\mathrm{1}} \theta\:{sin}^{\mathrm{2}{x}−\mathrm{1}} \theta\:{d}\theta \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} {cos}^{\mathrm{4}} \mathrm{3}\theta\:{sin}^{\mathrm{2}} \mathrm{6}\theta\:{d}\theta\:\overset{\theta\rightarrow\mathrm{3}\theta} {=}\:\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{4}} \theta\:{sin}^{\mathrm{2}} \mathrm{2}\theta\:{d}\theta\: \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{6}} \theta\:{sin}^{\mathrm{2}} \theta\:{d}\theta=\frac{\mathrm{2}}{\mathrm{3}}\beta\left(\frac{\mathrm{7}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{5}\right)}=\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\Gamma\left(\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{5}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\left(\frac{\mathrm{5}!!}{\mathrm{2}^{\mathrm{3}} }\sqrt{\pi}\:\right)\left(\frac{\mathrm{1}!!}{\mathrm{2}^{\mathrm{1}} }\:\sqrt{\pi}\:\right)}{\mathrm{4}!}=\frac{\mathrm{2}\pi}{\mathrm{3}}\:\frac{\left(\frac{\mathrm{15}}{\mathrm{8}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{24}} \\ $$$${I}=\frac{\mathrm{5}\pi}{\mathrm{192}} \\ $$

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