0-pi-2-dx-sin-x-

Question Number 42945 by ajfour last updated on 05/Sep/18

$$\int_{\mathrm{0}} ^{\:\:\pi/\mathrm{2}} \frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}}}\:=\:? \\ $$

Commented by MJS last updated on 05/Sep/18

this can′t be solved with elementar methods, it leads to an elliptic integral, you can find info on the web

$$\mathrm{this}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{with}\:\mathrm{elementar}\:\mathrm{methods}, \\ $$$$\mathrm{it}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{an}\:\mathrm{elliptic}\:\mathrm{integral},\:\mathrm{you}\:\mathrm{can}\:\mathrm{find} \\ $$$$\mathrm{info}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web} \\ $$

Commented by tannhattnnt28 last updated on 05/Sep/18

$$\mathrm{not} \\ $$

Commented by MJS last updated on 05/Sep/18

$$\mathrm{not}??? \\ $$

Commented by ajfour last updated on 05/Sep/18

$${if}\:{it}\:{is}\:{within}\:{indicated}\:{limits}\:? \\ $$

Commented by ajfour last updated on 05/Sep/18

Commented by ajfour last updated on 05/Sep/18

If a rod pivoted at one end be released in horizontal position, find the time it takes to swing down; (assume frictionless bearings).

$${If}\:{a}\:{rod}\:{pivoted}\:{at}\:{one}\:{end}\:{be} \\ $$$${released}\:{in}\:{horizontal}\:{position}, \\ $$$${find}\:{the}\:{time}\:{it}\:{takes}\:{to}\:{swing} \\ $$$${down};\:\left({assume}\:{frictionless}\right. \\ $$$$\left.{bearings}\right). \\ $$

Commented by ajfour last updated on 05/Sep/18

$${here}\:{is}\:{where}\:{i}\:{encounter}\:{above} \\ $$$${integral},\:{if}\:{i}\:{am}\:{not}\:{mislead}.. \\ $$

Commented by maxmathsup by imad last updated on 05/Sep/18

changement tan((x/2))=t give I = ∫_0 ^1 (1/( (√((2t)/(1+t^2 ))))) ((2dt)/(1+t^2 )) = ∫_0 ^1 ((2dt)/( (√(2t))(√(1+t^2 ))))dt =(√2)∫_0 ^1 (dt/( (√t)(√(1+t^2 )))) =_(t =sh(u)) (√2)∫_0 ^(ln(1+(√2))) ((ch(u)du)/( (√(sh(u)))ch(u))) =(√2) ∫_0 ^(ln(1+(√2))) (du/( (√(shu)))) ...i think this integral can t besolved by elementary function and if you have a try post it...

$${changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\:\sqrt{\mathrm{2}{t}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{{t}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$=_{{t}\:={sh}\left({u}\right)} \:\:\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\:\frac{{ch}\left({u}\right){du}}{\:\sqrt{{sh}\left({u}\right)}{ch}\left({u}\right)}\:=\sqrt{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\frac{{du}}{\:\sqrt{{shu}}}\:…{i}\:{think}\:{this}\:{integral} \\ $$$${can}\:{t}\:{besolved}\:{by}\:{elementary}\:{function}\:{and}\:{if}\:{you}\:{have}\:{a}\:{try}\:{post}\:{it}… \\ $$$$ \\ $$

Commented by ajfour last updated on 05/Sep/18

α=(τ/I) = ((mg(l/2)cos θ)/((ml^2 )/3)) = ((3gcos θ)/(2l)) ∫_0 ^( 𝛚) 𝛚d𝛚 = ((3g)/(2l))sin 𝛉 ⇒ 𝛚 =(d𝛉/dt) = (√((3gsin 𝛉)/l)) ⇒ t = (√((l/(3g)) ))∫_0 ^( 𝛑/2) (d𝛉/( (√(sin 𝛉)))) t ≈ 2.64(√(l/(3g))) .

$$\alpha=\frac{\tau}{{I}}\:=\:\frac{{mg}\frac{{l}}{\mathrm{2}}\mathrm{cos}\:\theta}{\frac{{ml}^{\mathrm{2}} }{\mathrm{3}}}\:=\:\frac{\mathrm{3}{g}\mathrm{cos}\:\theta}{\mathrm{2}{l}} \\ $$$$\int_{\mathrm{0}} ^{\:\:\boldsymbol{\omega}} \boldsymbol{\omega{d}\omega}\:=\:\frac{\mathrm{3}\boldsymbol{{g}}}{\mathrm{2}\boldsymbol{{l}}}\mathrm{sin}\:\boldsymbol{\theta} \\ $$$$\Rightarrow\:\:\boldsymbol{\omega}\:=\frac{\boldsymbol{{d}\theta}}{\boldsymbol{{dt}}}\:=\:\sqrt{\frac{\mathrm{3}\boldsymbol{{g}}\mathrm{sin}\:\boldsymbol{\theta}}{\boldsymbol{{l}}}} \\ $$$$\Rightarrow\:\:\boldsymbol{{t}}\:=\:\sqrt{\frac{\boldsymbol{{l}}}{\mathrm{3}\boldsymbol{{g}}}\:}\int_{\mathrm{0}} ^{\:\:\boldsymbol{\pi}/\mathrm{2}} \frac{\boldsymbol{{d}\theta}}{\:\sqrt{\mathrm{sin}\:\boldsymbol{\theta}}}\: \\ $$$$\:{t}\:\approx\:\mathrm{2}.\mathrm{64}\sqrt{\frac{{l}}{\mathrm{3}{g}}}\:. \\ $$

Commented by maxmathsup by imad last updated on 05/Sep/18

changement tan((x/2))=t give ∫_0 ^1 (1/( (√((2t)/(1+t^2 ))))) ((2dt)/(1+t^2 )) = 2 ∫_0 ^1 (dt/( (√(2t))(√(1+t^2 )))) =(√2)∫_0 ^1 (dt/( (√t)(√(1+t^2 )))) but we have (1+u)^α =1+(α/(1!)) u +((α(α−1))/(2!)) u^2 + ((α(α−1)....(α−n+1))/(n!)) u^n +...⇒ (√(1+t^2 ))=(1+t^2 )^(−(1/2)) =1−(1/2)t^2 +(((−(1/2))(−(1/2)−1))/2) t^4 +... =1−(t^2 /2) +(3/8) t^4 +...⇒(1/( (√t)(√(1+t^2 )))) = t^(−(1/2)) {1−(t^2 /2) +(3/8)t^4 +....} =t^(−(1/2)) −(t^(3/2) /2) +(3/8) t^(7/2) +....⇒∫_0 ^1 (dt/( (√t)(√(1+t^2 )))) = ∫_0 ^1 t^(−(1/2)) dt −(1/2) ∫_0 ^1 t^(3/2) dt +(3/8) ∫_0 ^1 t^(7/2) dt+... =[2(√t)]_0 ^1 −(1/2)[ (1/((3/2)+1)) t^((3/2)+1) ]_0 ^1 +(3/8)[ (1/((7/2)+1)) t^((7/2)+1) ]_0 ^1 −... =2 −(1/2) (2/5) +(3/8) (2/9) +....=2−(1/5) + (1/(12)) −...⇒ ∫_0 ^(π/2) (dx/( (√(sinx)))) = (√2){ 2−(1/5) +(1/(12)) +....} ∼ 2(√2) −((√2)/5) +((√2)/(12)) .

$${changement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{\mathrm{2}{t}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\:\sqrt{{t}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:{but}\:{we}\:{have} \\ $$$$\left(\mathrm{1}+{u}\right)^{\alpha} \:=\mathrm{1}+\frac{\alpha}{\mathrm{1}!}\:{u}\:\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}\:{u}^{\mathrm{2}} \:\:+\:\frac{\alpha\left(\alpha−\mathrm{1}\right)….\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}\:{u}^{{n}} \:+…\Rightarrow \\ $$$$\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \:+\frac{\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}}\:{t}^{\mathrm{4}} \:+… \\ $$$$=\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:\:+\frac{\mathrm{3}}{\mathrm{8}}\:{t}^{\mathrm{4}} \:+…\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{t}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\:{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left\{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}{t}^{\mathrm{4}} \:+….\right\} \\ $$$$={t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:−\frac{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:{t}^{\frac{\mathrm{7}}{\mathrm{2}}} \:+….\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\:\sqrt{{t}}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\frac{\mathrm{3}}{\mathrm{2}}} {dt}\:+\frac{\mathrm{3}}{\mathrm{8}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\frac{\mathrm{7}}{\mathrm{2}}} \:{dt}+… \\ $$$$=\left[\mathrm{2}\sqrt{{t}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:−\frac{\mathrm{1}}{\mathrm{2}}\left[\:\frac{\mathrm{1}}{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}}\:{t}^{\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:+\frac{\mathrm{3}}{\mathrm{8}}\left[\:\frac{\mathrm{1}}{\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{1}}\:{t}^{\frac{\mathrm{7}}{\mathrm{2}}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−… \\ $$$$=\mathrm{2}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{5}}\:+\frac{\mathrm{3}}{\mathrm{8}}\:\frac{\mathrm{2}}{\mathrm{9}}\:+….=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{12}}\:−…\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\:\sqrt{{sinx}}}\:=\:\sqrt{\mathrm{2}}\left\{\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{5}}\:+\frac{\mathrm{1}}{\mathrm{12}}\:+….\right\}\:\sim\:\mathrm{2}\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}\:. \\ $$

Commented by maxmathsup by imad last updated on 05/Sep/18

(1/( (√(1+t^2 )))) =(1+t^2 )^(−(1/2)) =....

$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}\:=\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=…. \\ $$

Commented by maxmathsup by imad last updated on 05/Sep/18

2(√2) −((√2)/5) +((√2)/(12)) ∼2×1,41 −((1,41)/5) +((1,41)/(12)) = 2,65 ⇒ ∫_0 ^(π/2) (dx/( (√(sinx)))) ∼ 2,65 .

$$\mathrm{2}\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{5}}\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{12}}\:\sim\mathrm{2}×\mathrm{1},\mathrm{41}\:−\frac{\mathrm{1},\mathrm{41}}{\mathrm{5}}\:+\frac{\mathrm{1},\mathrm{41}}{\mathrm{12}}\:=\:\mathrm{2},\mathrm{65}\:\Rightarrow\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\:\sqrt{{sinx}}}\:\sim\:\mathrm{2},\mathrm{65}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Sep/18

using gamma beta function ∫_0 ^(Π/2) sin^((−1)/2) x cos^0 x dx ∫_0 ^(Π/2) sin^(2p−1) xcos^(2q−1) xdx=((⌈(p)⌈(q))/(2⌈(p+q))) here 2p−1=((−1)/2) 2p=(1/2) p=(1/4) 2q−1=0 q=(1/2) so ans is =((⌈((1/4)) ×⌈((1/2)))/(2⌈((1/4)+(1/2)))) =((⌈((1/4)) ×⌈((1/2)))/(2⌈((3/4)))) =((⌈((1/4)) ×⌈((1/4)) ×(√Π))/(2×⌈((1/4))×⌈(1−(1/4)))) formula ⌈(p)⌈(1−p)=(Π/(sin(pΠ))) ⌈((1/4))×⌈(1−(1/4))=(Π/(sin((Π/4))))=Π×(√2) =((⌈((1/4))×⌈((1/4))×(√Π))/(2×Π×(√2)))= =((3.6×3.6×(√(3.14)))/(2×1.41×3.14))≈2.64

$${using}\:{gamma}\:{beta}\:{function} \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\frac{−\mathrm{1}}{\mathrm{2}}} {x}\:{cos}^{\mathrm{0}} {x}\:\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} {xcos}^{\mathrm{2}{q}−\mathrm{1}} {xdx}=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\mathrm{2}\lceil\left({p}+{q}\right)} \\ $$$${here}\:\mathrm{2}{p}−\mathrm{1}=\frac{−\mathrm{1}}{\mathrm{2}}\:\:\:\:\mathrm{2}{p}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:{p}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{2}{q}−\mathrm{1}=\mathrm{0}\:\:{q}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${so}\:{ans}\:{is}\:=\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:×\lceil\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\lceil\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$=\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:×\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:×\sqrt{\Pi}}{\mathrm{2}×\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)×\lceil\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$${formula}\:\lceil\left({p}\right)\lceil\left(\mathrm{1}−{p}\right)=\frac{\Pi}{{sin}\left({p}\Pi\right)} \\ $$$$\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)×\lceil\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\Pi}{{sin}\left(\frac{\Pi}{\mathrm{4}}\right)}=\Pi×\sqrt{\mathrm{2}}\: \\ $$$$=\frac{\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)×\lceil\left(\frac{\mathrm{1}}{\mathrm{4}}\right)×\sqrt{\Pi}}{\mathrm{2}×\Pi×\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{3}.\mathrm{6}×\mathrm{3}.\mathrm{6}×\sqrt{\mathrm{3}.\mathrm{14}}}{\mathrm{2}×\mathrm{1}.\mathrm{41}×\mathrm{3}.\mathrm{14}}\approx\mathrm{2}.\mathrm{64} \\ $$

Commented by ajfour last updated on 05/Sep/18

$${let}\:{me}\:{take}\:{some}\:{time}\:{to}\:{recap} \\ $$$${beta}\:{and}\:{gamma}\:{functions}\:{and} \\ $$$${see}\:{if}\:{i}\:{can}\:{understand}\:{your} \\ $$$${solution};\:{thanks}\:{a}\:{lot}\:{Sir}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Sep/18

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Sep/18

Commented by ajfour last updated on 05/Sep/18

$${excellent}\:{sir},\:{thanks}\:{again}! \\ $$

Commented by MJS last updated on 06/Sep/18

found something else: Γ^2 ((1/4))=2(√(2π))ϖ ⇒ ∫_0 ^(π/2) (dx/( (√(sin x))))=ϖ ϖ=(π/(M(1, (√2))))=the lemniscate constant a_(n+1) =((a_n +b_n )/2) b_(n+1) =(√(a_n b_n )) lim_(n→∞) a_n =lim_(n→∞) b_n =M(a_0 , b_0 )=the arithmetic−geometric mean of a_0 and b_0 both converge very fast, after 3 steps the mistake is only ≈1.7×10^(−8) M(1; (√2))≈1.198140235 ϖ=(π/(M(1, (√2))))≈2.622057554 btw.: ∫_0 ^(π/2) (√(sin x))dx=((2π)/ϖ)≈2.396280469

$$\mathrm{found}\:\mathrm{something}\:\mathrm{else}: \\ $$$$\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{2}\sqrt{\mathrm{2}\pi}\varpi\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}}}=\varpi \\ $$$$\varpi=\frac{\pi}{{M}\left(\mathrm{1},\:\sqrt{\mathrm{2}}\right)}=\mathrm{the}\:\mathrm{lemniscate}\:\mathrm{constant} \\ $$$$ \\ $$$${a}_{{n}+\mathrm{1}} =\frac{{a}_{{n}} +{b}_{{n}} }{\mathrm{2}} \\ $$$${b}_{{n}+\mathrm{1}} =\sqrt{{a}_{{n}} {b}_{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}{b}_{{n}} ={M}\left({a}_{\mathrm{0}} ,\:{b}_{\mathrm{0}} \right)=\mathrm{the}\:\mathrm{arithmetic}−\mathrm{geometric}\:\mathrm{mean}\:\mathrm{of}\:{a}_{\mathrm{0}} \:\mathrm{and}\:{b}_{\mathrm{0}} \\ $$$$\mathrm{both}\:\mathrm{converge}\:\mathrm{very}\:\mathrm{fast},\:\mathrm{after}\:\mathrm{3}\:\mathrm{steps}\:\mathrm{the} \\ $$$$\mathrm{mistake}\:\mathrm{is}\:\mathrm{only}\:\approx\mathrm{1}.\mathrm{7}×\mathrm{10}^{−\mathrm{8}} \\ $$$${M}\left(\mathrm{1};\:\sqrt{\mathrm{2}}\right)\approx\mathrm{1}.\mathrm{198140235} \\ $$$$\varpi=\frac{\pi}{{M}\left(\mathrm{1},\:\sqrt{\mathrm{2}}\right)}\approx\mathrm{2}.\mathrm{622057554} \\ $$$$ \\ $$$$\mathrm{btw}.: \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\sqrt{\mathrm{sin}\:{x}}{dx}=\frac{\mathrm{2}\pi}{\varpi}\approx\mathrm{2}.\mathrm{396280469} \\ $$

Commented by ajfour last updated on 06/Sep/18

$${great}\:{hunting}\:{Sir},\:{thanks}\:{for} \\ $$$${sharing}. \\ $$

Commented by Necxx last updated on 19/Sep/18

$${hmmm} \\ $$

Answered by MJS last updated on 05/Sep/18

∫_0 ^(π/2) (dx/( (√(sin x))))≈2.62206 approximately calculated by my TI−89 calculator

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{sin}\:{x}}}\approx\mathrm{2}.\mathrm{62206} \\ $$$$\mathrm{approximately}\:\mathrm{calculated}\:\mathrm{by}\:\mathrm{my}\:\mathrm{TI}−\mathrm{89}\:\mathrm{calculator} \\ $$

Commented by MJS last updated on 05/Sep/18

his answer is right but it′s not easy to calculate the values of Γ(x) too. my calculator says, these values are questionable so if you have got a table of these values, good. I haven′t got one

$$\mathrm{his}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{right}\:\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{easy}\:\mathrm{to} \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\Gamma\left({x}\right)\:\mathrm{too}.\:\mathrm{my}\:\mathrm{calculator} \\ $$$$\mathrm{says},\:\mathrm{these}\:\mathrm{values}\:\mathrm{are}\:\mathrm{questionable} \\ $$$$\mathrm{so}\:\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{got}\:\mathrm{a}\:\mathrm{table}\:\mathrm{of}\:\mathrm{these}\:\mathrm{values},\:\mathrm{good}. \\ $$$$\mathrm{I}\:\mathrm{haven}'\mathrm{t}\:\mathrm{got}\:\mathrm{one} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Sep/18