0-1-arcsin-x-x-dx-

Question Number 156616 by amin96 last updated on 13/Oct/21

\int_{0}^{1} \frac{a r c s i n (x)}{x} d x = ?

Answered by mindispower last updated on 13/Oct/21

b y p a r t = - \int_{0}^{1} \frac{l n (x)}{\sqrt{1 - x^{2}}}

x^{2} = u = - \int_{0}^{1} \frac{l n (u)}{2 \sqrt{1 - u}} . \frac{d u}{2 \sqrt{u}}

= - \frac{1}{4} \int_{0}^{1} l n (u) u^{- \frac{1}{2}} {(1 - u)}^{- \frac{1}{2}} d u

= - \frac{1}{4} \partial_{a} \int_{0}^{1} u^{a} {(1 - u)}^{- \frac{1}{2}} d u ∣_{= \frac{1}{2}}

= - \frac{1}{4} \partial_{a} β (a, \frac{1}{2}) ∣_{a = \frac{1}{2_{}}}

= - \frac{1}{4} (Ψ (\frac{1}{2}) - Ψ (1)) β (\frac{1}{2}, \frac{1}{2})

Answered by puissant last updated on 13/Oct/21

Q = \int_{0}^{1} \frac{a r c s i n x}{x} d x

I B P \Rightarrow Q = - \int_{0}^{1} \frac{l n x}{\sqrt{1 - x^{2}}} d x

x = s i n u \to d x = c o s u d u

\Rightarrow Q = - \int_{0}^{\frac{π}{2}} \frac{l n (s i n u)}{c o s u} c o s u d u

\Rightarrow Q = - \int_{0}^{\frac{π}{2}} l n (s i n u) d u

\int_{0}^{\frac{π}{2}} l n (s i n x) d x = - \frac{π}{2} l n 2

∴∵ Q = \int_{0}^{1} \frac{a r c s i n x}{x} d x = \frac{π}{2} l n 2 . .

Commented by amin96 last updated on 13/Oct/21

\int_{0}^{\frac{π}{2}} l n (s i n (x)) d x \overset{? ? ? ?}{=} - \frac{π}{2} l n 2

Commented by puissant last updated on 13/Oct/21

y e s s s s s s s s s i r . .