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

∫_0 ^( π/2) (dx/( (√(sin x)))) = ?

\int_{0}^{π / 2} \frac{d x}{\sqrt{\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

this {can}^{'} t be solved with elementar methods,

it leads to an elliptic integral, you can find

info on the web

Commented by tannhattnnt28 last updated on 05/Sep/18

not

not

Commented by MJS last updated on 05/Sep/18

not???

not ? ? ?

Commented by ajfour last updated on 05/Sep/18

if it is within indicated limits ?

i f i t i s w i t h i n i n d i c a t e d l i m i t s ?

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).

I f a r o d p i v o t e d a t o n e e n d b e

r e l e a s e d i n h o r i z o n t a l p o s i t i o n,

f i n d t h e t i m e i t t a k e s t o s w i n g

d o w n; (a s s u m e f r i c t i o n l e s s

b e a r i n g s) .

Commented by ajfour last updated on 05/Sep/18

here is where i encounter above integral, if i am not mislead..

h e r e i s w h e r e i e n c o u n t e r a b o v e

i n t e g r a l, i f i a m n o t m i s l e a d . .

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...

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int_{0}^{1} \frac{1}{\sqrt{\frac{2 t}{1 + t^{2}}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{1} \frac{2 d t}{\sqrt{2 t} \sqrt{1 + t^{2}}} d t = \sqrt{2} \int_{0}^{1} \frac{d t}{\sqrt{t} \sqrt{1 + t^{2}}}

=_{t = s h (u)} \sqrt{2} \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (u) d u}{\sqrt{s h (u)} c h (u)} = \sqrt{2} \int_{0}^{l n (1 + \sqrt{2})} \frac{d u}{\sqrt{s h u}} \dots i t h i n k t h i s i n t e g r a l

c a n t b e s o l v e d b y e l e m e n t a r y f u n c t i o n a n d i f y o u h a v e a t r y p o s t i t \dots

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))) .

α = \frac{τ}{I} = \frac{m g \frac{l}{2} \cos θ}{\frac{{m l}^{2}}{3}} = \frac{3 g \cos θ}{2 l}

\int_{0}^{ω} ω d ω = \frac{3 g}{2 l} \sin θ

\Rightarrow ω = \frac{d θ}{d t} = \sqrt{\frac{3 g \sin θ}{l}}

\Rightarrow t = \sqrt{\frac{l}{3 g}} \int_{0}^{π / 2} \frac{d θ}{\sqrt{\sin θ}}

t \approx 2.64 \sqrt{\frac{l}{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)) .

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e \int_{0}^{1} \frac{1}{\sqrt{\frac{2 t}{1 + t^{2}}}} \frac{2 d t}{1 + t^{2}}

= 2 \int_{0}^{1} \frac{d t}{\sqrt{2 t} \sqrt{1 + t^{2}}} = \sqrt{2} \int_{0}^{1} \frac{d t}{\sqrt{t} \sqrt{1 + t^{2}}} b u t w e h a v e

{(1 + u)}^{α} = 1 + \frac{α}{1!} u + \frac{α (α - 1)}{2!} u^{2} + \frac{α (α - 1) \dots . (α - n + 1)}{n!} u^{n} + \dots \Rightarrow

\sqrt{1 + t^{2}} = {(1 + t^{2})}^{- \frac{1}{2}} = 1 - \frac{1}{2} t^{2} + \frac{(- \frac{1}{2}) (- \frac{1}{2} - 1)}{2} t^{4} + \dots

= 1 - \frac{t^{2}}{2} + \frac{3}{8} t^{4} + \dots \Rightarrow \frac{1}{\sqrt{t} \sqrt{1 + t^{2}}} = t^{- \frac{1}{2}} {1 - \frac{t^{2}}{2} + \frac{3}{8} t^{4} + \dots .}

= t^{- \frac{1}{2}} - \frac{t^{\frac{3}{2}}}{2} + \frac{3}{8} t^{\frac{7}{2}} + \dots . \Rightarrow \int_{0}^{1} \frac{d t}{\sqrt{t} \sqrt{1 + t^{2}}} = \int_{0}^{1} t^{- \frac{1}{2}} d t - \frac{1}{2} \int_{0}^{1} t^{\frac{3}{2}} d t + \frac{3}{8} \int_{0}^{1} t^{\frac{7}{2}} d t + \dots

= {[2 \sqrt{t}]}_{0}^{1} - \frac{1}{2} {[\frac{1}{\frac{3}{2} + 1} t^{\frac{3}{2} + 1}]}_{0}^{1} + \frac{3}{8} {[\frac{1}{\frac{7}{2} + 1} t^{\frac{7}{2} + 1}]}_{0}^{1} - \dots

= 2 - \frac{1}{2} \frac{2}{5} + \frac{3}{8} \frac{2}{9} + \dots . = 2 - \frac{1}{5} + \frac{1}{12} - \dots \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{s i n x}} = \sqrt{2} {2 - \frac{1}{5} + \frac{1}{12} + \dots .} \sim 2 \sqrt{2} - \frac{\sqrt{2}}{5} + \frac{\sqrt{2}}{12} .

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

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

\frac{1}{\sqrt{1 + t^{2}}} = {(1 + t^{2})}^{- \frac{1}{2}} = \dots .

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 .

2 \sqrt{2} - \frac{\sqrt{2}}{5} + \frac{\sqrt{2}}{12} \sim 2 \times 1, 41 - \frac{1, 41}{5} + \frac{1, 41}{12} = 2, 65 \Rightarrow \int_{0}^{\frac{π}{2}} \frac{d x}{\sqrt{s i n x}} \sim 2, 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

u s i n g g a m m a b e t a f u n c t i o n

\int_{0}^{\frac{Π}{2}} {s i n}^{\frac{- 1}{2}} x {c o s}^{0} x d x

\int_{0}^{\frac{Π}{2}} {s i n}^{2 p - 1} {x c o s}^{2 q - 1} x d x = \frac{⌈ (p) ⌈ (q)}{2 ⌈ (p + q)}

h e r e 2 p - 1 = \frac{- 1}{2} 2 p = \frac{1}{2} p = \frac{1}{4}

2 q - 1 = 0 q = \frac{1}{2}

s o a n s i s = \frac{⌈ (\frac{1}{4}) \times ⌈ (\frac{1}{2})}{2 ⌈ (\frac{1}{4} + \frac{1}{2})}

= \frac{⌈ (\frac{1}{4}) \times ⌈ (\frac{1}{2})}{2 ⌈ (\frac{3}{4})}

= \frac{⌈ (\frac{1}{4}) \times ⌈ (\frac{1}{4}) \times \sqrt{Π}}{2 \times ⌈ (\frac{1}{4}) \times ⌈ (1 - \frac{1}{4})}

f o r m u l a ⌈ (p) ⌈ (1 - p) = \frac{Π}{s i n (p Π)}

⌈ (\frac{1}{4}) \times ⌈ (1 - \frac{1}{4}) = \frac{Π}{s i n (\frac{Π}{4})} = Π \times \sqrt{2}

= \frac{⌈ (\frac{1}{4}) \times ⌈ (\frac{1}{4}) \times \sqrt{Π}}{2 \times Π \times \sqrt{2}} =

= \frac{3.6 \times 3.6 \times \sqrt{3.14}}{2 \times 1.41 \times 3.14} \approx 2.64

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.

l e t m e t a k e s o m e t i m e t o r e c a p

b e t a a n d g a m m a f u n c t i o n s a n d

s e e i f i c a n u n d e r s t a n d y o u r

s o l u t i o n; t h a n k s a l o t S i r .

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!

e x c e l l e n t s i r, t h a n k s a g a i n!

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

found something else :

Γ^{2} (\frac{1}{4}) = 2 \sqrt{2 π} ϖ \Rightarrow \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\sqrt{\sin x}} = ϖ

ϖ = \frac{π}{M (1, \sqrt{2})} = the lemniscate constant

a_{n + 1} = \frac{a_{n} + b_{n}}{2}

b_{n + 1} = \sqrt{a_{n} b_{n}}

\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} 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 \approx 1.7 \times 10^{- 8}

M (1; \sqrt{2}) \approx 1.198140235

ϖ = \frac{π}{M (1, \sqrt{2})} \approx 2.622057554

btw . :

\underset{0}{\int^{\frac{π}{2}}} \sqrt{\sin x} d x = \frac{2 π}{ϖ} \approx 2.396280469

Commented by ajfour last updated on 06/Sep/18

great hunting Sir, thanks for sharing.

g r e a t h u n t i n g S i r, t h a n k s f o r

s h a r i n g .

Commented by Necxx last updated on 19/Sep/18

hmmm

h m m m

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{0}{\int^{\frac{π}{2}}} \frac{d x}{\sqrt{\sin x}} \approx 2.62206

approximately calculated by my TI - 89 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

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

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

sir your answer is right sir...

s i r y o u r a n s w e r i s r i g h t s i r \dots

Commented by malwaan last updated on 06/Sep/18

TI−89 Is it apk ?

TI - 89

Is it apk ?

Commented by MJS last updated on 06/Sep/18

it is a device

it is a device

Commented by Joel578 last updated on 06/Sep/18

scientific calculator

scientific calculator

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