find ∫_0 ^1 ((sinx)/(1+x^2 ))dx


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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

$t r y s i r t o f i n d i t . . .$
Commented by abdo mathsup 649 cc last updated on 01/Jun/19
$another way we have x−(x^3 /6) ≤sinx ≤x ⇒ (x/(1+x^2 )) −(x^3 /(6(1+x^2 ))) ≤ ((sinx)/(1+x^2 )) ≤ (x/(1+x^2 )) ⇒ ∫_0 ^1 (x/(1+x^2 ))dx −(1/6) ∫_0 ^1 (x^3 /(1+x^2 ))dx ≤ ∫_0 ^1 ((sinx)/(1+x^2 ))dx≤∫_0 ^1 (x/(1+x^2 ))dx ∫_0 ^1 (x/(1+x^2 ))dx =[(1/2)ln(1+x^2 )]_0 ^1 =((ln(2))/2) ∫_0 ^1 (x^3 /(x^2 +1)) dx =∫_0 ^1 ((x(x^2 +1)−x)/(x^2 +1))dx =∫_0 ^1 xdx−∫_0 ^1 ((xdx)/(x^2 +1)) =[(x^2 /2)]_0 ^1 −((ln(2))/2) =(1/2) −((ln(2))/2) ⇒ ((ln(2))/2) −(1/(12)) +((ln(2))/(12)) ≤ ∫_0 ^1 ((sinxdx)/(1+x^2 )) ≤ ((ln(2))/2) ⇒ ((7ln(2)−1)/(12)) ≤ ∫_0 ^1 ((sinx)/(1+x^2 ))dx ≤ ((ln(2))/2) so we can take v_0 =((7ln(2)−1)/(24)) +((ln(2{)/4) =((13ln(2)−1)/(24)) as approximst value for this integral .$
$a n o t h e r w a y w e h a v e x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x \Rightarrow$ $\frac{x}{1 + x^{2}} - \frac{x^{3}}{6 (1 + x^{2})} ⩽ \frac{s i n x}{1 + x^{2}} ⩽ \frac{x}{1 + x^{2}} \Rightarrow$ $\int_{0}^{1} \frac{x}{1 + x^{2}} d x - \frac{1}{6} \int_{0}^{1} \frac{x^{3}}{1 + x^{2}} d x ⩽ \int_{0}^{1} \frac{s i n x}{1 + x^{2}} d x ⩽ \int_{0}^{1} \frac{x}{1 + x^{2}} d x$ $\int_{0}^{1} \frac{x}{1 + x^{2}} d x = {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{1} = \frac{l n (2)}{2}$ $\int_{0}^{1} \frac{x^{3}}{x^{2} + 1} d x = \int_{0}^{1} \frac{x (x^{2} + 1) - x}{x^{2} + 1} d x$ $= \int_{0}^{1} x d x - \int_{0}^{1} \frac{x d x}{x^{2} + 1} = {[\frac{x^{2}}{2}]}_{0}^{1} - \frac{l n (2)}{2} = \frac{1}{2} - \frac{l n (2)}{2} \Rightarrow$ $\frac{l n (2)}{2} - \frac{1}{12} + \frac{l n (2)}{12} ⩽ \int_{0}^{1} \frac{s i n x d x}{1 + x^{2}} ⩽ \frac{l n (2)}{2} \Rightarrow$ $\frac{7 l n (2) - 1}{12} ⩽ \int_{0}^{1} \frac{s i n x}{1 + x^{2}} d x ⩽ \frac{l n (2)}{2} s o w e c a n t a k e$ $v_{0} = \frac{7 l n (2) - 1}{24} + \frac{l n (2 {}{4} = \frac{13 l n (2) - 1}{24} a s a p p r o x i m s t$ $v a l u e f o r t h i s i n t e g r a l .$
Commented by abdo mathsup 649 cc last updated on 01/Jun/19

$l e t I = \int_{0}^{1} \frac{s i n x}{1 + x^{2}} d x w e h a v e s i n x = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} x^{2 n + 1}}{(2 n + 1)!} \Rightarrow$ $I = \int_{0}^{1} (\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} x^{2 n + 1} \frac{1}{1 + x^{2}}) d x$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} \int_{0}^{1} \frac{x^{2 n + 1}}{1 + x^{2}} d x l e t w_{n} = \int_{0}^{1} \frac{x^{2 n + 1}}{1 + x^{2}} d x$ $\Rightarrow w_{n} =_{x = t a n θ} \int_{0}^{\frac{π}{4}} \frac{{t a n}^{2 n + 1} θ (1 + {t a n}^{2} θ) d θ}{1 + {t a n}^{2} θ}$ $= \int_{0}^{\frac{π}{4}} {t a n}^{2 n + 1} θ d θ (t h e v a l u e o f t h i s i n t e g r a l i s$ $k n o w n) \Rightarrow I = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{(2 n + 1)!} w_{n}$
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