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

Question Number 173148 by JordanRoddy last updated on 07/Jul/22

\int_{0}^{1}^{n} \sqrt{x} (\arcsin x) dx

Answered by Mathspace last updated on 07/Jul/22

I_{n} = \int_{0}^{1} (^{n} \sqrt{x}) a r c s i n x d x w e d o

t h e c h a n g e m e n t a r c s i n x = t \Rightarrow

x = s i n t a n d

I_{n} = \int_{0}^{\frac{π}{2}} {(s i n t)}^{\frac{1}{n}} t c o s t d t

= \int_{0}^{\frac{π}{2}} t (c o s t {(s i n t)}^{\frac{1}{n}}) d t

= {[\frac{t}{1 + \frac{1}{n}} {(s i n t)}^{1 + \frac{1}{n}}]}_{0}^{\frac{π}{2}}

- \int_{0}^{\frac{π}{2}} \frac{n}{n + 1} {(s i n t)}^{1 + \frac{1}{n}} d t

\frac{π n}{2 (n + 1)} - \frac{n}{n + 1} \int_{0}^{\frac{π}{2}} {(s i n t)}^{1 + \frac{1}{n}} d t

w e h a v e \int_{0}^{\frac{π}{2}} {(c o s t)}^{2 p - 1} . {(s i n t)}^{2 q - 1} d t

= B (ρ, q) = \frac{Γ (ρ) . Γ (q)}{Γ (ρ + q)} \Rightarrow

2 ρ - 1 = 0 a n d 2 q - 1 = 1 + \frac{1}{n}

\Rightarrow ρ = \frac{1}{2} a n d q = 1 + \frac{1}{2 n}

\int_{0}^{\frac{π}{2}} {(s i n t)}^{1 + \frac{1}{n}} d t = \frac{Γ (\frac{1}{2}) Γ (1 + \frac{1}{2 n})}{2 Γ (\frac{1}{2} + 1 + \frac{1}{2 n})}

= \frac{\sqrt{π}}{2} \times \frac{Γ (1 + \frac{1}{2 n})}{Γ (\frac{3}{2} + \frac{1}{2 n})} \Rightarrow

I_{n} = \frac{n π}{2 (n + 1)} - \frac{n}{n + 1} . \frac{\sqrt{π}}{2} \times \frac{Γ (1 + \frac{1}{2 n})}{Γ (\frac{3}{2} + \frac{1}{2 n})}

Commented by Mathspace last updated on 07/Jul/22

s o r r y . . \int_{0}^{\frac{π}{2}} {(c o s t)}^{2 ρ - 1} {(s i n t)}^{2 q - 1} d t

= \frac{1}{2} B (p, q) = \frac{Γ (ρ) . Γ (q)}{2 Γ (ρ + q)}

Commented by Tawa11 last updated on 11/Jul/22

Great sir

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