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Question Number 222635    Answers: 0   Comments: 0

Question Number 222622    Answers: 0   Comments: 0

Question Number 222619    Answers: 2   Comments: 0

Question Number 222614    Answers: 0   Comments: 0

complicated solution of the integral ; ∫ tan(((cos x)/x))^(13) dx

$$ \\ $$$$\:\:\mathrm{complicated}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral}\:; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\mathrm{tan}\left(\frac{\mathrm{cos}\:{x}}{{x}}\right)^{\mathrm{13}} \:\mathrm{d}{x} \\ $$$$ \\ $$

Question Number 222613    Answers: 0   Comments: 0

what is I = ∫ tan (((cos x)/x)) dx

$$ \\ $$$$\:\:\:\:\:\:\mathrm{what}\:\mathrm{is}\:\:{I}\:=\:\int\:\mathrm{tan}\:\left(\frac{\mathrm{cos}\:{x}}{{x}}\right)\:\mathrm{d}{x}\: \\ $$$$ \\ $$

Question Number 222616    Answers: 1   Comments: 0

Question Number 222610    Answers: 0   Comments: 0

Σ_(n=1) ^∞ (1/(n(e^(2πn) − 1)))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({e}^{\mathrm{2}\pi{n}} \:−\:\mathrm{1}\right)} \\ $$$$ \\ $$

Question Number 222624    Answers: 1   Comments: 0

Question Number 222607    Answers: 1   Comments: 0

Prove:∫_0 ^1 (x^3 /(5−x^3 ))∙(1/( ((1−x^3 ))^(1/3) ))dx=((((10))^(1/3) −2)/(3(√3)))π

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} }{\mathrm{5}−{x}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−{x}^{\mathrm{3}} }}{dx}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{10}}−\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\pi \\ $$

Question Number 222601    Answers: 0   Comments: 1

L 60 1.

$$\mathrm{L}\:\mathrm{60} \\ $$$$ \\ $$$$\mathrm{1}.\:\: \\ $$

Question Number 222599    Answers: 0   Comments: 1

Σ_(k=1) ^∞ (−1)^(k+1) (1/((2k−1)^(2n+1) ))=(π^(2n+1) /(2^(2n+2) (2n)!))∣E_(2m) ∣

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$

Question Number 222592    Answers: 0   Comments: 0

Σ_(n = 1) ^∞ (1/(cosh^4 [n ln((√2) + 1)]))

$$\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{cosh}^{\mathrm{4}} \left[\mathrm{n}\:\mathrm{ln}\left(\sqrt{\mathrm{2}}\:\:+\:\:\mathrm{1}\right)\right]} \\ $$

Question Number 222585    Answers: 3   Comments: 2

Question Number 222584    Answers: 0   Comments: 0

find the correct const. to preconst. “ x^n −2=−x ”

$${find}\:{the}\:{correct}\:{const}.\:{to}\:{preconst}.\:``\:{x}^{{n}} −\mathrm{2}=−{x}\:'' \\ $$

Question Number 222582    Answers: 2   Comments: 0

If: a_i > 0 , b_i > 0 , i = 1,...,n^(−) Prove that: (√(a_1 ^2 + b_1 ^2 )) + (√(a_2 ^2 + b_2 ^2 )) +...+ (√(a_n ^2 + b_n ^2 )) ≥ ≥ (√((a_1 +a_2 +...+a_n )^2 + (b_1 +b_2 +...+b_n )^2 ))

$$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},...,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+...+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} }\:\:\:\geqslant \\ $$$$\geqslant\:\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +...+\mathrm{a}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} +...+\mathrm{b}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} } \\ $$

Question Number 222579    Answers: 1   Comments: 0

Question Number 222595    Answers: 0   Comments: 2

Ω=∫ ((sin^2 x (√(cos x)))/(1+3sin^2 x))dx I can′t solve this, maybe some genius can. WolframAlpha gives a solution but I need the path...

$$\Omega=\int\:\frac{\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:{x}}}{\mathrm{1}+\mathrm{3sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{this},\:\mathrm{maybe}\:\mathrm{some}\:\mathrm{genius}\:\mathrm{can}. \\ $$$$\mathrm{WolframAlpha}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{I} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{path}... \\ $$

Question Number 222577    Answers: 1   Comments: 0

Question Number 222567    Answers: 0   Comments: 0

Prove:∫_0 ^1 (√((((√(K^2 +36K′^2 ))+6K^′ )/(K^2 +36K^(′2) )) ))(dk/( (√k)(1−k^2 )^(2/3) ))=(√π)((√2)−(√((4−2(√2))/3))) ∫_0 ^1 (√(((√(K^2 +36K′^2 ))+6K^2 )/(K^2 +36K^(′2) ))) (dk/( (√k)(1−k^2 )^(2/3) )) k=sinθ⇒dk=cosθdθ,θ∈[0,(π/2)] (√k)=sinθ,(1−k^2 )^(2/3) =cos^(4/3) θ (dk/( (√k)(1−k^2 )^(2/3) ))=((cosθdθ)/( (√(sinθ))cos^(4/3) θ))=sin^(−(1/2)) θcos^(−(1/3)) θdθ K=K(sin θ),K′=K′(sinθ)=K(cosθ) ∫_0 ^(π/2) (√(((√(K^2 (sin θ)+36K′^2 (sin θ)))+6K′(sin θ))/(K^2 (sinθ)+36K′^2 (sinθ))))sin^(−(1/2)) θcos^(−(1/3)) θdθ K(k)=(π/2) _2 F_1 ((1/2),(1/2);1;k^2 ),K′(k)=(π/2) _2 F_1 ((1/2),(1/2);1;1−k^2 ) (√((√(K^2 +36′^2 ))+6K′))(√(K^2 +36K′^2 ))=(K/( (√((K^2 +36K′^2 )((√(K^2 +36K′^2 ))−6K′))))) (√(K^2 +36K^2 ))−6K′=(K^2 /( (√(K^2 +26K^(′2) ))+6K^′ )) (K/( (√((K^2 +36K′^2 )((√(K^2 +36K^′^2 ))−6K′)))))=(K/( (√(K^2 ((√(K^2 +36K′^2 +))6K^′ ))))) =(1/( (√((√(K^2 +36K^(′2) ))+6K′)))) ∫_0 ^(π/2) (1/( (√((√(K^2 (sin θ)+36K^2 (sin θ)))+o6K′(sin θ)))))sin^(−(1/2)) θ cos^(−(1/3)) θdθ K(sin θ)=(π/2)Σ_(n=0) ^∞ ((((1/2))_n ^2 )/((n!)^2 ))sin^(2n) θ,K′(sin θ)=(π/2)Σ_(n=0) ^∞ ((((1/2))_n ^2 )/((n!)^2 ))cos^(2n) θ (√(K^3 +36K^(′2) ))+6K′=(π/2)Σ_(n=0) ^∞ ((((1/2))_n ^2 )/((n!)^2 ))(sin^(2n) θ+36cos^(2n) θ+6∙cos^(2n) θ) =(π/2)Σ_(n=0) ^∞ ((((1/2))_n ^2 )/((n!)^2 ))(sin^(2n) θ+48 cos^(2n) θ) (1/( (√((√(K^2 +36K′^2 ))+6K′))))=Σ_(n=0) ^∞ (((n!)^2 )/(((1/2))_n ^2 )) (1/(sin^(2n) θ+48 cos^(2n) θ)) ∫_0 ^(π/2) (1/π)Σ_(n=0) ^∞ (((n!)^2 )/(((1/2))_n ^2 )) (1/(sin^(2n) θ+48cos^(2n) θ))sin^(−(1/2)) θ cos^(−(1/3)) θdθ =(2/π)Σ_(n=0) ^∞ (((n!)^2 )/(((1/2))_n ^2 ))∫_0 ^(π/2) ((sin^(−(1/2)) θ cos^(−(1/3)) θ)/(sin^(2n) θ+48cos^(2n) θ))dθ t=tan^2 θ⇒dθ=(dt/(2(√t)(1+t))),sin^2 θ=(t/(1+t)),cos^2 θ=(1/(1+t)) ∫_0 ^∞ ((((t/(1+t)))^(−(1/3)) )/(((t/(1+t)))^n +48((1/(1+t)))^n )) (dt/(2(√t)(1+t))) =∫_0 ^∞ ((t^(−(1/4)) (1+t)^(1/4) (1+t)^(1/6) )/(((t^n /((1−t)^n ))+48(1/((1+t)^n ))))) (dt/(2(√t)(1+t))) =∫_0 ^∞ ((t^(−(1/4)) (1+t)^((1/4)+(1/6)) )/((t^n +48)/((1+t)^n ))) (dt/(2t^(1/2) (1+t))) =(1/2) ∫_0 ^∞ ((t^(−(3/4)) (1+t)^((1/4)+(1/6)−n+1) )/(t^n +48))dt =(1/2)∫_0 ^∞ ((t^(−(3/4)) (1+t)^(((13)/(12))−n) )/(t^n +48))dt u=(t/(1+t))⇒t=(u/(1−u)),dt=(du/((1−u)^2 )) (1/2)∫_0 ^1 ((((u/(1−u)))^(−(3/4)) ((1/(1−u)))^(((13)/(12))−n) )/(((u/(1−u)))^n +48)) (du/((1−u)^2 )) =(1/2)∫_0 ^1 u^(−(3/4)) (1−u)^(3/4) (1−u)^(−((13)/(12))+12) (1−u)^2 (du/(((u^n /((1−u)^n ))+48))) =(1/2)∫_0 ^1 u^(−(3/4)) (1−u)^((3/4)−((13)/(12))n+2) (du/((u^n /((1−u)^n ))+48)) =(1/2)∫_0 ^1 u^(−(3/4)) (1−u)^((9/(12))−((13)/(12))+n+2) (((1−u)^u du)/(u^n +48(1−u)^n )) =(1/2)∫_0 ^1 u^(−(3/4)) (1−u)^(n+((23)/(12))) (du/(u^n +48(1−u)^n )) =(1/1)∫_0 ^1 u^(−(3/4)) (1−u)^(n+((23)/(12))) (du/(1+48(((1−u)/u))^n )) v=((1−u)/u)⇒u=(1/(1+v)),du=−(dv/((1+v)^2 )) (1/2)∫_∞ ^0 ((1/(1+v)))^(−(3/4)−n) v^(n+((23)/(12))) (1/(1+48v^n ))(−(dv/((1+v)^2 ))) =(1/2)∫_0 ^∞ (1+v)^((3/4)+n) v^(n+((23)/(12))) (1/(1+48v^n )) (dv/((1+v)^2 )) =(1/2)∫_0 ^∞ v^(n+((21)/(11))) (1+v)^((3/4)+n−2) (dv/(1+48v^u )) =(1/2)∫_0 ^∞ v^(n+((23)/(12))) (1+v)^(v−(5/4)) (dv/(1+48v^n )) w=e^n ⇒dv=(1/n)w^((1/n)−1) dw (1/2)∫_0 ^∞ w^((n+((23)/(12)))/n) (1+w^(1/n) )^(n−(5/4)) (1/(1+48w)) (1/n)w^((1/n)−1) dw =(1/(2n))∫_0 ^∞ w^((n/n)+((23)/(12n))+(1/n)−1) (1+w^(1/n) )^(n−(5/4)) (dw/(1+48w)) =(1/(2n))∫_0 ^∞ w^(1+((23)/(12n))+(1/n)−1) (1+w^(1/n) )^(n−(5/4)) (dw/(1+48w)) =(1/(2n))∫_0 ^∞ w^(((35)/(12n))+(1/n)) (1+w^(1/n) )^(n−(5/4)) (dw/(1+48w)) ≈(1/(2n))∫_0 ^∞ w^((35)/(12n)) e^((n−(5/4))w^(1/n) ) (dw/(1+48n)) for large n Γ(((35)/(12))+1)((Γ((1/2)))/(Γ(((35)/(12n))+(3/2)))) (1/(48^(((35)/(12n))+1) )) Σ_(n=0) ^∞ (((n!)^2 )/(((1/2))_n ^2 )) (1/(2n)) Γ(((35)/(12n))+1)((Γ((1/2)))/(Γ(((35)/(12n))+(3/2)))) (1/(48^(((35)/(12n))+1) ))=(√π)((√2)−(√((4−2(√2))/3)))

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{'} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }\:}\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{\mathrm{2}} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }}\:\frac{\mathrm{d}{k}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$${k}=\mathrm{sin}\theta\Rightarrow\mathrm{d}{k}=\mathrm{cos}\theta\mathrm{d}\theta,\theta\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right] \\ $$$$\sqrt{{k}}=\mathrm{sin}\theta,\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} =\mathrm{cos}^{\frac{\mathrm{4}}{\mathrm{3}}} \theta \\ $$$$\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{cos}\theta\mathrm{d}\theta}{\:\sqrt{\mathrm{sin}\theta}\mathrm{cos}^{\frac{\mathrm{4}}{\mathrm{3}}} \theta}=\mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{3}}} \theta\mathrm{d}\theta \\ $$$${K}={K}\left(\mathrm{sin}\:\theta\right),{K}'={K}'\left(\mathrm{sin}\theta\right)={K}\left(\mathrm{cos}\theta\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} \left(\mathrm{sin}\:\theta\right)+\mathrm{36}{K}'^{\mathrm{2}} \left(\mathrm{sin}\:\theta\right)}+\mathrm{6}{K}'\left(\mathrm{sin}\:\theta\right)}{{K}^{\mathrm{2}} \left(\mathrm{sin}\theta\right)+\mathrm{36}{K}'^{\mathrm{2}} \left(\mathrm{sin}\theta\right)}}\mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{3}}} \theta\mathrm{d}\theta \\ $$$${K}\left({k}\right)=\frac{\pi}{\mathrm{2}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};{k}^{\mathrm{2}} \right),{K}'\left({k}\right)=\frac{\pi}{\mathrm{2}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{1}−{k}^{\mathrm{2}} \right) \\ $$$$\sqrt{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}'^{\mathrm{2}} }+\mathrm{6}{K}'}\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }=\frac{{K}}{\:\sqrt{\left({K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} \right)\left(\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }−\mathrm{6}{K}'\right)}} \\ $$$$\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}^{\mathrm{2}} }−\mathrm{6}{K}'=\frac{{K}^{\mathrm{2}} }{\:\sqrt{{K}^{\mathrm{2}} +\mathrm{26}{K}^{'\mathrm{2}} }+\mathrm{6}{K}^{'} } \\ $$$$\frac{{K}}{\:\sqrt{\left({K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} \right)\left(\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'^{\mathrm{2}} } }−\mathrm{6}{K}'\right)}}=\frac{{K}}{\:\sqrt{{K}^{\mathrm{2}} \left(\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} +}\mathrm{6}{K}^{'} \right)}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }+\mathrm{6}{K}'}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\:\sqrt{\sqrt{{K}^{\mathrm{2}} \left(\mathrm{sin}\:\theta\right)+\mathrm{36}{K}^{\mathrm{2}} \left(\mathrm{sin}\:\theta\right)}+\mathrm{o6}{K}'\left(\mathrm{sin}\:\theta\right)}}\mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\:\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{3}}} \theta\mathrm{d}\theta \\ $$$${K}\left(\mathrm{sin}\:\theta\right)=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({n}!\right)^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}{n}} \theta,{K}'\left(\mathrm{sin}\:\theta\right)=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({n}!\right)^{\mathrm{2}} }\mathrm{cos}^{\mathrm{2}{n}} \theta \\ $$$$\sqrt{{K}^{\mathrm{3}} +\mathrm{36}{K}^{'\mathrm{2}} }+\mathrm{6}{K}'=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({n}!\right)^{\mathrm{2}} }\left(\mathrm{sin}^{\mathrm{2}{n}} \theta+\mathrm{36cos}^{\mathrm{2}{n}} \theta+\mathrm{6}\centerdot\mathrm{cos}^{\mathrm{2}{n}} \theta\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({n}!\right)^{\mathrm{2}} }\left(\mathrm{sin}^{\mathrm{2}{n}} \theta+\mathrm{48}\:\mathrm{cos}^{\mathrm{2}{n}} \theta\right) \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}'}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}{n}} \theta+\mathrm{48}\:\mathrm{cos}^{\mathrm{2}{n}} \theta}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\pi}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}{n}} \theta+\mathrm{48cos}^{\mathrm{2}{n}} \theta}\mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\:\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{3}}} \theta\mathrm{d}\theta \\ $$$$=\frac{\mathrm{2}}{\pi}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \theta\:\mathrm{cos}^{−\frac{\mathrm{1}}{\mathrm{3}}} \theta}{\mathrm{sin}^{\mathrm{2}{n}} \theta+\mathrm{48cos}^{\mathrm{2}{n}} \theta}\mathrm{d}\theta \\ $$$${t}=\mathrm{tan}^{\mathrm{2}} \theta\Rightarrow\mathrm{d}\theta=\frac{{dt}}{\mathrm{2}\sqrt{{t}}\left(\mathrm{1}+{t}\right)},\mathrm{sin}^{\mathrm{2}} \theta=\frac{{t}}{\mathrm{1}+{t}},\mathrm{cos}^{\mathrm{2}} \theta=\frac{\mathrm{1}}{\mathrm{1}+{t}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\left(\frac{{t}}{\mathrm{1}+{t}}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} }{\left(\frac{{t}}{\mathrm{1}+{t}}\right)^{{n}} +\mathrm{48}\left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right)^{{n}} }\:\frac{{dt}}{\mathrm{2}\sqrt{{t}}\left(\mathrm{1}+{t}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} }{\left(\frac{{t}^{{n}} }{\left(\mathrm{1}−{t}\right)^{{n}} }+\mathrm{48}\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{{n}} }\right)}\:\frac{{dt}}{\mathrm{2}\sqrt{{t}}\left(\mathrm{1}+{t}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}} }{\frac{{t}^{{n}} +\mathrm{48}}{\left(\mathrm{1}+{t}\right)^{{n}} }}\:\frac{{dt}}{\mathrm{2}{t}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{t}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{6}}−{n}+\mathrm{1}} }{{t}^{{n}} +\mathrm{48}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{t}\right)^{\frac{\mathrm{13}}{\mathrm{12}}−{n}} }{{t}^{{n}} +\mathrm{48}}{dt} \\ $$$${u}=\frac{{t}}{\mathrm{1}+{t}}\Rightarrow{t}=\frac{{u}}{\mathrm{1}−{u}},{dt}=\frac{{du}}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\frac{{u}}{\mathrm{1}−{u}}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)^{\frac{\mathrm{13}}{\mathrm{12}}−{n}} }{\left(\frac{{u}}{\mathrm{1}−{u}}\right)^{{n}} +\mathrm{48}}\:\frac{{du}}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{13}}{\mathrm{12}}+\mathrm{12}} \left(\mathrm{1}−{u}\right)^{\mathrm{2}} \frac{{du}}{\left(\frac{{u}^{{n}} }{\left(\mathrm{1}−{u}\right)^{{n}} }+\mathrm{48}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{13}}{\mathrm{12}}{n}+\mathrm{2}} \frac{{du}}{\frac{{u}^{{n}} }{\left(\mathrm{1}−{u}\right)^{{n}} }+\mathrm{48}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{9}}{\mathrm{12}}−\frac{\mathrm{13}}{\mathrm{12}}+{n}+\mathrm{2}} \frac{\left(\mathrm{1}−{u}\right)^{{u}} {du}}{{u}^{{n}} +\mathrm{48}\left(\mathrm{1}−{u}\right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{{n}+\frac{\mathrm{23}}{\mathrm{12}}} \frac{{du}}{{u}^{{n}} +\mathrm{48}\left(\mathrm{1}−{u}\right)^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{{n}+\frac{\mathrm{23}}{\mathrm{12}}} \frac{{du}}{\mathrm{1}+\mathrm{48}\left(\frac{\mathrm{1}−{u}}{{u}}\right)^{{n}} } \\ $$$${v}=\frac{\mathrm{1}−{u}}{{u}}\Rightarrow{u}=\frac{\mathrm{1}}{\mathrm{1}+{v}},{du}=−\frac{{dv}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\infty} ^{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{1}+{v}}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}−{n}} {v}^{{n}+\frac{\mathrm{23}}{\mathrm{12}}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{48}{v}^{{n}} }\left(−\frac{{dv}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \left(\mathrm{1}+{v}\right)^{\frac{\mathrm{3}}{\mathrm{4}}+{n}} {v}^{{n}+\frac{\mathrm{23}}{\mathrm{12}}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{48}{v}^{{n}} }\:\frac{{dv}}{\left(\mathrm{1}+{v}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {v}^{{n}+\frac{\mathrm{21}}{\mathrm{11}}} \left(\mathrm{1}+{v}\right)^{\frac{\mathrm{3}}{\mathrm{4}}+{n}−\mathrm{2}} \frac{{dv}}{\mathrm{1}+\mathrm{48}{v}^{{u}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {v}^{{n}+\frac{\mathrm{23}}{\mathrm{12}}} \left(\mathrm{1}+{v}\right)^{{v}−\frac{\mathrm{5}}{\mathrm{4}}} \frac{{dv}}{\mathrm{1}+\mathrm{48}{v}^{{n}} } \\ $$$${w}={e}^{{n}} \Rightarrow{dv}=\frac{\mathrm{1}}{{n}}{w}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dw} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {w}^{\frac{{n}+\frac{\mathrm{23}}{\mathrm{12}}}{{n}}} \left(\mathrm{1}+{w}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}−\frac{\mathrm{5}}{\mathrm{4}}} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{48}{w}}\:\frac{\mathrm{1}}{{n}}{w}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dw} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\infty} {w}^{\frac{{n}}{{n}}+\frac{\mathrm{23}}{\mathrm{12}{n}}+\frac{\mathrm{1}}{{n}}−\mathrm{1}} \left(\mathrm{1}+{w}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}−\frac{\mathrm{5}}{\mathrm{4}}} \frac{{dw}}{\mathrm{1}+\mathrm{48}{w}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\infty} {w}^{\mathrm{1}+\frac{\mathrm{23}}{\mathrm{12}{n}}+\frac{\mathrm{1}}{{n}}−\mathrm{1}} \left(\mathrm{1}+{w}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}−\frac{\mathrm{5}}{\mathrm{4}}} \frac{{dw}}{\mathrm{1}+\mathrm{48}{w}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\infty} {w}^{\frac{\mathrm{35}}{\mathrm{12}{n}}+\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}+{w}^{\frac{\mathrm{1}}{{n}}} \right)^{{n}−\frac{\mathrm{5}}{\mathrm{4}}} \frac{{dw}}{\mathrm{1}+\mathrm{48}{w}} \\ $$$$\approx\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\infty} {w}^{\frac{\mathrm{35}}{\mathrm{12}{n}}} {e}^{\left({n}−\frac{\mathrm{5}}{\mathrm{4}}\right){w}^{\frac{\mathrm{1}}{{n}}} } \frac{{dw}}{\mathrm{1}+\mathrm{48}{n}}\:\mathrm{for}\:\mathrm{large}\:{n} \\ $$$$\Gamma\left(\frac{\mathrm{35}}{\mathrm{12}}+\mathrm{1}\right)\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{35}}{\mathrm{12}{n}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\frac{\mathrm{1}}{\mathrm{48}^{\frac{\mathrm{35}}{\mathrm{12}{n}}+\mathrm{1}} } \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }\:\frac{\mathrm{1}}{\mathrm{2}{n}}\:\Gamma\left(\frac{\mathrm{35}}{\mathrm{12}{n}}+\mathrm{1}\right)\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{35}}{\mathrm{12}{n}}+\frac{\mathrm{3}}{\mathrm{2}}\right)}\:\frac{\mathrm{1}}{\mathrm{48}^{\frac{\mathrm{35}}{\mathrm{12}{n}}+\mathrm{1}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$

Question Number 222546    Answers: 3   Comments: 2

Question Number 222533    Answers: 4   Comments: 0

Question Number 222532    Answers: 0   Comments: 0

Prove: Σ_(n=0) ^∞ (1/2^(4n) ) (((2n)),(n) )^2 (K_(n+1) /((2n+1)^2 ))=((192)/π)ILi_4 (((1+i)/2))+((16)/π)ILi_3 (((1+i)/2))ln(2)+((15π^2 )/8)ln(2)+(1/2)ln(2)^3 −((148)/π)β(4),K_n =1+(1/3)+…+(1/(2n−1))

$$\mathrm{Prove}: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}{n}} }\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \frac{\mathscr{K}_{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{192}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{4}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)+\frac{\mathrm{16}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{15}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)^{\mathrm{3}} −\frac{\mathrm{148}}{\pi}\beta\left(\mathrm{4}\right),\mathscr{K}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\ldots+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$

Question Number 222531    Answers: 2   Comments: 0

The foot of the perpendicular from a point of the circle x^2 +y^2 =1,z=0 to the plan 2x+3y+z=6 lie on curve−−−−−−

$${The}\:{foot}\:{of}\:{the}\:{perpendicular}\:{from}\: \\ $$$${a}\:{point}\:{of}\:{the}\:{circle}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1},{z}=\mathrm{0} \\ $$$${to}\:{the}\:{plan}\:\mathrm{2}{x}+\mathrm{3}{y}+{z}=\mathrm{6}\:{lie}\:{on}\:{curve}−−−−−− \\ $$

Question Number 222529    Answers: 0   Comments: 0

∫_0 ^∞ x(x^2 +1)^(−x+(1/x) ) dx = ??

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−{x}+\frac{\mathrm{1}}{{x}}\:} \:{dx}\:=\:?? \\ $$$$ \\ $$

Question Number 222552    Answers: 1   Comments: 0

x^x =64 , x=?

$${x}^{{x}} =\mathrm{64}\:\:\:\:\:\:,\:\:{x}=? \\ $$

Question Number 222523    Answers: 2   Comments: 0

x+V−J(x)((ustx)/(zac^2 x))=x−((ustx_0 )/( (√x)ψ+ζ(((−1+v%)/(π+2))+L_(l(x→0)) ^( v%) X_(2x^2 ) ^( 1) )))

$${x}+{V}−{J}\left({x}\right)\frac{{ustx}}{{zac}^{\mathrm{2}} {x}}={x}−\frac{{ustx}_{\mathrm{0}} }{\:\sqrt{{x}}\psi+\zeta\left(\frac{−\mathrm{1}+{v\%}}{\pi+\mathrm{2}}+\mathscr{L}_{{l}\left({x}\rightarrow\mathrm{0}\right)} ^{\:\:{v\%}} {X}_{\mathrm{2}{x}^{\mathrm{2}} } ^{\:\mathrm{1}} \right)} \\ $$

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