0-1-sinx-dx-

Question Number 187453 by norboyev last updated on 17/Feb/23

\underset{0}{\int^{1}} \sqrt{\sin x} d x = ?

Answered by Frix last updated on 17/Feb/23

\int \sqrt{\sin x} d x = [Elliptic Integral]

= - 2 E (\frac{π}{4} - \frac{x}{2} ∣ 2) + C

\underset{0}{\int^{1}} \sqrt{\sin x} d x \approx . 642977634658

Answered by CElcedricjunior last updated on 21/Feb/23

\int_{0}^{1} \sqrt{s i n x} d x = k

p o s o n s s i n x = t^{2} => d x = \frac{2 t}{\sqrt{1 - t^{2}}} d t

q d : {\begin{cases} x - > 0 \\ x - > 1 \end{cases} => {\begin{cases} t - > 0 \\ t - > s i n (1) \end{cases}

k = \int_{0}^{s i n (1)} \frac{2 t^{3}}{\sqrt{1 - t^{2}}} d t

{\begin{cases} u = t^{2} \\ v^{'} = \frac{t}{\sqrt{1 - t^{2}}} \end{cases} => {\begin{cases} u^{'} = 2 t \\ v = - \sqrt{1 - t^{2}} \end{cases} ★ M o i v r e

k = - 2 s i \overset{2}{n} (1) ∣ c o s (1) ∣ + 4 \int_{0}^{s i n (1)} t \sqrt{1 - t^{2}} d t

k = - s i n (1) s i n (2) + 4 {[- \frac{2}{3} \sqrt{(1 - t^{2})}]}_{0}^{s i n (1)} C e d r i c j u n i o r

k = s i n (1) s i n (2) - \frac{8}{3} c o \overset{\frac{3}{2}}{s} (1) + \frac{8}{3}