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Question Number 130844 by mnjuly1970 last updated on 29/Jan/21
... advanced calculus... prove that:: ∫_0 ^( (π/3)) (dx/( ((cos^2 (x)))^(1/3) )) =((2^(1/3) (√π))/( (√3)))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:\:\:{advanced}\:\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{3}}} \frac{{dx}}{\:\sqrt[{\mathrm{3}}]{{cos}^{\mathrm{2}} \left({x}\right)}}\:=\frac{\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{3}}} \sqrt{\pi}}{\:\sqrt{\mathrm{3}}} \\ $$
Answered by Dwaipayan Shikari last updated on 29/Jan/21
∫_0 ^(π/3) (cosx)^(−(2/3)) dx=−∫_1 ^(1/2) t^(−(2/3)) (1−t^2 )^(−(1/2)) dt =∫_0 ^1 t^(−(2/3)) (1−t^2 )^(−(1/2)) dt−∫_0 ^(1/2) t^(−(5/6)) (1−t^2 )^(−(1/2)) dt =(1/2)∫_0 ^1 u^(−(5/6)) (1−u)^(−(1/2)) du−(1/2)∫_0 ^(1/( (√2))) u^(−(5/6)) (1−u)^(−(1/2)) du =((Γ((( 1)/6))Γ((1/2)))/(2Γ((2/3))))−(1/2)Σ_(n=0) ^∞ ((((1/2))_n )/(n!))∫_0 ^(1/( (√2))) u^(n−(5/6)) du =((Γ((5/6))(√π))/(2Γ((2/3))))−(3/( (2)^(1/3) )) _2 F_1 ((1/6),(1/2),(7/6),(1/4))
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \left({cosx}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} {dx}=−\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt}−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {t}^{−\frac{\mathrm{5}}{\mathrm{6}}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{5}}{\mathrm{6}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {u}^{−\frac{\mathrm{5}}{\mathrm{6}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$=\frac{\Gamma\left(\frac{\:\mathrm{1}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} {u}^{{n}−\frac{\mathrm{5}}{\mathrm{6}}} {du} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\sqrt{\pi}}{\mathrm{2}\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}−\frac{\mathrm{3}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{6}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{6}},\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$

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