find-0-1-sinx-1-x-2-dx-

Question Number 61326 by maxmathsup by imad last updated on 01/Jun/19

f i n d \int_{0}^{1} \frac{s i n x}{1 + x^{2}} d x

Commented by perlman last updated on 01/Jun/19

e x a c t e v a l u e!

Commented by maxmathsup by imad last updated on 01/Jun/19

t r y s i r t o f i n d i t \dots

Commented by abdo mathsup 649 cc last updated on 01/Jun/19

a n o t h e r w a y w e h a v e x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x \Rightarrow

\frac{x}{1 + x^{2}} - \frac{x^{3}}{6 (1 + x^{2})} ⩽ \frac{s i n x}{1 + x^{2}} ⩽ \frac{x}{1 + x^{2}} \Rightarrow

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

\int_{0}^{1} \frac{x}{1 + x^{2}} d x = {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{1} = \frac{l n (2)}{2}

\int_{0}^{1} \frac{x^{3}}{x^{2} + 1} d x = \int_{0}^{1} \frac{x (x^{2} + 1) - x}{x^{2} + 1} d x

= \int_{0}^{1} x d x - \int_{0}^{1} \frac{x d x}{x^{2} + 1} = {[\frac{x^{2}}{2}]}_{0}^{1} - \frac{l n (2)}{2} = \frac{1}{2} - \frac{l n (2)}{2} \Rightarrow

\frac{l n (2)}{2} - \frac{1}{12} + \frac{l n (2)}{12} ⩽ \int_{0}^{1} \frac{s i n x d x}{1 + x^{2}} ⩽ \frac{l n (2)}{2} \Rightarrow

\frac{7 l n (2) - 1}{12} ⩽ \int_{0}^{1} \frac{s i n x}{1 + x^{2}} d x ⩽ \frac{l n (2)}{2} s o w e c a n t a k e

v_{0} = \frac{7 l n (2) - 1}{24} + \frac{l n (2 {}{4} = \frac{13 l n (2) - 1}{24} a s a p p r o x i m s t

v a l u e f o r t h i s i n t e g r a l .

Commented by abdo mathsup 649 cc last updated on 01/Jun/19

l e t I = \int_{0}^{1} \frac{s i n x}{1 + x^{2}} d x w e h a v e s i n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1)!} \Rightarrow

I = \int_{0}^{1} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1} \frac{1}{1 + x^{2}}) d x

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} \int_{0}^{1} \frac{x^{2 n + 1}}{1 + x^{2}} d x l e t w_{n} = \int_{0}^{1} \frac{x^{2 n + 1}}{1 + x^{2}} d x

\Rightarrow w_{n} =_{x = t a n θ} \int_{0}^{\frac{π}{4}} \frac{{t a n}^{2 n + 1} θ (1 + {t a n}^{2} θ) d θ}{1 + {t a n}^{2} θ}

= \int_{0}^{\frac{π}{4}} {t a n}^{2 n + 1} θ d θ (t h e v a l u e o f t h i s i n t e g r a l i s

k n o w n) \Rightarrow I = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} w_{n}