prove-that-0-pi-x-sin-x-1-2-dx-pi-8-2-1-4-

Question Number 141914 by mnjuly1970 last updated on 24/May/21

p r o v e t h a t ::

ϕ := \int_{0}^{π} \frac{x}{{(s i n (x))}^{\frac{1}{2}}} d x = \sqrt{\frac{π}{8}} Γ^{2} (\frac{1}{4}) \dots ✓

\dots \dots \dots

Answered by qaz last updated on 24/May/21

ϕ = \int_{0}^{π} \frac{x}{\sqrt{\sin x}} dx \dots \dots \dots \dots \dots \dots . . (1)

= \int_{0}^{π} \frac{π - x}{\sqrt{\sin x}} dx

= π \int_{0}^{π} \frac{dx}{\sqrt{\sin x}} - \int_{0}^{π} \frac{x}{\sqrt{\sin x}} dx \dots \dots . (2)

(1) + (2)

\Rightarrow ϕ = \frac{π}{2} \int_{0}^{π} \frac{dx}{\sqrt{\sin x}} = π \int_{0}^{π / 2} \frac{dx}{\sqrt{\sin x}}

= \frac{π}{2} \frac{Γ (\frac{1}{4}) Γ (\frac{1}{2})}{Γ (\frac{3}{4})}

= \frac{π \sqrt{π}}{2} \frac{Γ^{2} (\frac{1}{4})}{π} \sin \frac{π}{4}

= \sqrt{\frac{π}{8}} Γ^{2} (\frac{1}{4})

Answered by Dwaipayan Shikari last updated on 24/May/21

\int_{0}^{π} \frac{x}{\sqrt{s i n x}} d x = \int_{0}^{π} \frac{π - x}{\sqrt{s i n x}} d x = \frac{π}{2} \int_{0}^{π} \frac{1}{\sqrt{s i n (x)}} d x = π . \frac{Γ (\frac{1}{4}) Γ (\frac{1}{2})}{2 Γ (\frac{3}{4})}

= \sqrt{\frac{π}{2}} \frac{Γ^{2} (\frac{1}{4})}{2} = \sqrt{\frac{π}{8}} Γ^{2} (\frac{1}{4})

Answered by mathmax by abdo last updated on 24/May/21

Φ = \int_{0}^{π} \frac{x}{\sqrt{sinx}} dx \Rightarrow Φ =_{x = π - t} - \int_{0}^{π} \frac{π - t}{\sqrt{sint}} (- dt) = \int_{0}^{π} \frac{π dt}{\sqrt{sint}} - Φ \Rightarrow

2 Φ = π \int_{0}^{π} \frac{dt}{\sqrt{sint}} \Rightarrow Φ = \frac{π}{2} \int_{0}^{π} \frac{dt}{\sqrt{sint}}

we have \int_{0}^{π} \frac{dt}{\sqrt{sint}} = \int_{0}^{\frac{π}{2}} \frac{dt}{\sqrt{sint}} + \int_{\frac{π}{2}}^{π} \frac{dt}{\sqrt{sint}} (\to t = \frac{π}{2} + u)

= \int_{0}^{\frac{π}{2}} \frac{dt}{\sqrt{sint}} + \int_{0}^{\frac{π}{2}} \frac{du}{\sqrt{cosu}} but

\int_{0}^{\frac{π}{2}} \frac{dt}{\sqrt{sint}} = \int_{0}^{\frac{π}{2}} {(cost)}^{0} {(sint)}^{- \frac{1}{2}} dt

2 p - 1 = 0 \Rightarrow p = \frac{1}{2} and 2 q - 1 = - \frac{1}{2} \Rightarrow 2 q = \frac{1}{2} \Rightarrow q = \frac{1}{4} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{dt}{\sqrt{sint}} = \int_{0}^{\frac{π}{2}} {(cost)}^{2 . \frac{1}{2} - 1} . {(sint)}^{2 . \frac{1}{4} - 1} dt = \frac{1}{2} B (\frac{1}{2}, \frac{1}{4})

= \frac{1}{2} \frac{Γ (\frac{1}{2}) . Γ (\frac{1}{4})}{Γ (\frac{3}{4})}

Γ (\frac{1}{4}) . Γ (1 - \frac{1}{4}) = \frac{π}{\sin (\frac{π}{4})} = π \sqrt{2} \Rightarrow Γ (\frac{3}{4}) = \frac{π \sqrt{2}}{Γ (\frac{1}{4})} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{dt}{\sqrt{sint}} = \frac{\sqrt{π}}{2} \times π \sqrt{2} . Γ^{2} (\frac{1}{4}) = \frac{π \sqrt{π}}{\sqrt{2}} Γ^{2} (\frac{1}{4}) also

\int_{0}^{\frac{π}{2}} \frac{dt}{\sqrt{cost}} = \frac{π \sqrt{π}}{\sqrt{2}} Γ^{2} (\frac{1}{4}) \Rightarrow \int_{0}^{π} \frac{dt}{\sqrt{sint}} = \frac{2 π \sqrt{π}}{\sqrt{2}} Γ^{2} (\frac{1}{4}) \Rightarrow

Φ = π \sqrt{2 π} . Γ^{2} (\frac{1}{4})