Question Number 208176 by mnjuly1970 last updated on 07/Jun/24
$$ \\ $$$$\:\:\:\:\:\boldsymbol{{Find}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{the}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{folloing}}\:\boldsymbol{{integral}}. \\ $$$$\:\:\:\:\:\:\: \\ $$$$\begin{array}{|c|}{\:\:\:\boldsymbol{\Omega}=\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\:\mathrm{1}}{\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\:\boldsymbol{{cosx}}}}\:\boldsymbol{{dx}}\:=\:?\:\:}\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\: \\ $$
Commented by Frix last updated on 07/Jun/24
$$\mathrm{See}\:\mathrm{question}\:\mathrm{208140} \\ $$
Answered by Berbere last updated on 07/Jun/24
![(1/(1+((cos(x)))^(1/3) ))=Σ(−1)^n cos^(n/3) (x) Ω=Σ_(n≥0) ∫_0 ^(π/2) (−1)^n cos^(2((n/6)+(1/2))−1) (x)sin^(2((1/2))−1) (x)dx =Σ_(n≥0) (((−1)^n )/2)β((n/6)+(1/2),(1/2))=Σ_(n≥0) (((−1)^n )/2)((Γ((n/6)+(1/2))Γ((1/2)))/(Γ((n/6)+1))) N=∪_(k∈{0,1,2,3,4,5)) (6N+k) =Σ_(k=0) ^5 Σ_(m=0) ^∞ (((−1)^(6m+k) )/2).((Γ(m+(k/6)+(1/2)))/(Γ(m+(k/6)+1))) Γ((1/2)) =Γ((1/2))Σ_(k=0) ^5 (((−1)^k )/2)Σ_(m≥0) (((Γ(m+(k/6)+(1/2)))/(Γ(m)))/((Γ(m+(k/6)+1))/(Γ(m)))).Γ(m+1).(1^m /(m!)) =((√π)/2)Σ_(k=0) ^5 _2 F_1 ((k/6)+(1/2),1;(k/6)+1;[1])](https://www.tinkutara.com/question/Q208181.png)
$$\frac{\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{cos}\left({x}\right)}}=\Sigma\left(−\mathrm{1}\right)^{{n}} {cos}^{\frac{{n}}{\mathrm{3}}} \left({x}\right) \\ $$$$\Omega=\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−\mathrm{1}\right)^{{n}} {cos}^{\mathrm{2}\left(\frac{{n}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \left({x}\right){sin}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\mathrm{1}} \left({x}\right){dx} \\ $$$$=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\beta\left(\frac{{n}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}\frac{\Gamma\left(\frac{{n}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{n}}{\mathrm{6}}+\mathrm{1}\right)} \\ $$$$\mathbb{N}=\underset{{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right)} {\cup}\left(\mathrm{6}\mathbb{N}+{k}\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{6}{m}+{k}} }{\mathrm{2}}.\frac{\Gamma\left({m}+\frac{{k}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({m}+\frac{{k}}{\mathrm{6}}+\mathrm{1}\right)}\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}}\underset{{m}\geqslant\mathrm{0}} {\sum}\frac{\frac{\Gamma\left({m}+\frac{{k}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({m}\right)}}{\frac{\Gamma\left({m}+\frac{{k}}{\mathrm{6}}+\mathrm{1}\right)}{\Gamma\left({m}\right)}}.\Gamma\left({m}+\mathrm{1}\right).\frac{\mathrm{1}^{{m}} }{{m}!} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{{k}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1};\frac{{k}}{\mathrm{6}}+\mathrm{1};\left[\mathrm{1}\right]\right) \\ $$
Commented by Frix last updated on 07/Jun/24
$$\mathrm{Nice}! \\ $$
Commented by Berbere last updated on 10/Jun/24
$${thank}\:{You}\:{sir}\:{Have}\:\:{a}\:{nice}\:{Day} \\ $$