is-it-possible-to-find-the-exact-value-of-I-I-0-pi-sin-sin-x-dx-

Question Number 57653 by MJS last updated on 09/Apr/19

is it possible to find the exact value of I ?

I = \underset{0}{\int^{π}} \sin (\sin x) d x

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

Commented by maxmathsup by imad last updated on 10/Apr/19

i t s o n l y a t r y I = \int_{0}^{\frac{π}{2}} s i n (s i n x) d x + \int_{\frac{π}{2}}^{π} s i n (s i n x) d x

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

= \int_{0}^{1} \frac{s i n t}{\sqrt{1 - t^{2}}} d t l e t φ (t) = s i n t w e h a v e φ (t) = φ (0) + t φ^{^{'}} (0) + \frac{t^{2}}{2} φ^{(2)} (0) + \frac{t^{3}}{3!} φ^{(3)} (0) + \dots

φ^{^{'}} (t) = c o s t \Rightarrow φ^{^{'}} (0) = 1

φ^{(2)} (t) = - s i n t \Rightarrow φ^{(2)} (0) = 0

φ^{(3)} (t) = - c o s t \Rightarrow φ^{(3)} (0) = - 1 \Rightarrow φ (t) = t - \frac{t^{3}}{6} + 0 (t^{3}) \Rightarrow

t - \frac{t^{3}}{6} ⩽ s i n t ⩽ t f o r t \in [0, \frac{π}{2}] \Rightarrow \frac{t - \frac{t^{3}}{6}}{\sqrt{1 - t^{2}}} ⩽ \frac{s i n t}{\sqrt{1 - t^{2}}} ⩽ \frac{t}{\sqrt{1 - t^{2}}} \Rightarrow

\int_{0}^{1} \frac{t}{\sqrt{1 - t^{2}}} - \frac{1}{6} \int_{0}^{1} \frac{t^{3}}{\sqrt{1 - t^{2}}} d t ⩽ \int_{0}^{1} \frac{s i n t}{\sqrt{1 - t^{2}}} d t ⩽ \int_{0}^{1} \frac{t d t}{\sqrt{1 - t^{2}}}

\int_{0}^{1} \frac{t}{\sqrt{1 - t^{2}}} d t = {[- \sqrt{1 - t^{2}}]}_{0}^{1} = 1

\int_{0}^{1} \frac{t^{3}}{\sqrt{1 - t^{2}}} d t =_{t = s i n θ} \int_{0}^{\frac{π}{2}} \frac{{s i n}^{3} θ}{c o s θ} c o s θ d θ = \int_{0}^{\frac{π}{2}} {s i n}^{3} θ d θ

= \int_{0}^{\frac{π}{2}} s i n θ (1 - {c o s}^{2} θ) d θ = \int_{0}^{\frac{π}{2}} s i n θ d θ - \int_{0}^{\frac{π}{2}} s i n θ {c o s}^{2} θ d θ = {[- c o s θ]}_{0}^{\frac{π}{2}} + {[\frac{1}{3} {c o s}^{3} θ]}_{0}^{\frac{π}{2}}

= 1 - \frac{1}{3} = \frac{2}{3} \Rightarrow 1 - \frac{1}{6} \frac{2}{3} ⩽ \int_{0}^{1} \frac{s i n t}{\sqrt{1 - t^{2}}} d t ⩽ 1 \Rightarrow \frac{8}{9} ⩽ \int_{0}^{1} \frac{s i n t}{\sqrt{1 - t^{2}}} ⩽ 1 a l s o

\int_{\frac{π}{2}}^{π} s i n (s i n x d x =_{x = \frac{π}{2} + t} \int_{0}^{\frac{π}{2}} s i n (c o s t) d t =_{t = \frac{π}{2} - u} - \int_{0}^{\frac{π}{2}} s i n (s i n u) (- d u)

= \int_{0}^{\frac{π}{2}} s i n (s i n u) d u \Rightarrow I = 2 \int_{0}^{\frac{π}{2}} s i n (s i n u) d u = 2 \int_{0}^{1} \frac{s i n x}{\sqrt{1 - x^{2}}} d x \Rightarrow

\frac{16}{9} ⩽ \int_{0}^{π} s i n (s i n x) d x ⩽ 2 w e c a n t a k e v_{0} = \frac{2 + \frac{16}{9}}{2} = 1 + \frac{16}{18} = 1 + \frac{8}{9} = \frac{17}{9} \sim 1, 8

a s a p p r o x i m a t e v a l u e .

d e s m o s g i v e \int \sim 1, 78 a n d 1, 78 \sim 1, 8 s o t h i s a n s w e r i s a b e t t e r v a l u e \dots .

Commented by MJS last updated on 10/Apr/19

thank you for trying

Commented by Abdo msup. last updated on 10/Apr/19

y o u a r e a l w a y s w e l c o m e s i r .