let f(a) =∫_0 ^1 ((sin(2x))/(1+ax^2 )) dx with ∣a∣<1 1) approximate f(a) by a polynom 2) find the value (perhaps not exact) of ∫_0 ^1 ((sin(2x))/(1+2x^2 )) dx 3) let g(a) = ∫_0 ^1 ((x^2 sin(2x))/((1+ax^2 )^2 )) dx approximat g(a) by a polynom 4) find the value of ∫


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Question Number 61328 by maxmathsup by imad last updated on 01/Jun/19

$l e t f (a) = \int_{0}^{1} \frac{s i n (2 x)}{1 + {a x}^{2}} d x w i t h ∣ a ∣< 1$ $1) a p p r o x i m a t e f (a) b y a p o l y n o m$ $2) f i n d t h e v a l u e (p e r h a p s n o t e x a c t) o f \int_{0}^{1} \frac{s i n (2 x)}{1 + 2 x^{2}} d x$ $3) l e t g (a) = \int_{0}^{1} \frac{x^{2} s i n (2 x)}{{(1 + {a x}^{2})}^{2}} d x a p p r o x i m a t g (a) b y a p o l y n o m$ $4) f i n d t h e v a l u e o f \int_{0}^{1} \frac{x^{2} s i n (2 x)}{{(1 + 2 x^{2})}^{2}} d x .$
Commented bymaxmathsup by imad last updated on 02/Jun/19

$1) t h e Q i s a p p r o x i m a t e f (a) b y a f u n c t i o n .$
Commented bymaxmathsup by imad last updated on 02/Jun/19
$1) we have x−(x^3 /6) ≤ sinx ≤ x ⇒2x−((8x^3 )/6) ≤ sin(2x) ≤2x ⇒ 2x−(4/3)x^3 ≤ sin(2x)≤2x ⇒((2x−(4/3)x^3 )/(1+ax^2 )) ≤((sin(2x))/(1+ax^2 )) ≤ ((2x)/(1+ax^2 )) ⇒ ∫_0 ^1 ((2xdx)/(1+ax^2 )) −(4/3) ∫_0 ^1 (x^3 /(1+ax^2 ))dx ≤ ∫_0 ^1 ((sin(2x))/(1+ax^2 )) dx ≤ ∫_0 ^1 ((2x)/(1+ax^2 )) dx ∫_0 ^1 ((2xdx)/(ax^2 +1)) =(1/a) ∫_0 ^1 ((2ax dx)/(ax^2 +1)) =(1/a)[ln(ax^2 +1)]_0 ^1 =((ln(1+a))/a) ( we suppose a≠0 ) ∫_0 ^1 (x^3 /(ax^2 +1)) dx =(1/a) ∫_0 ^1 (((ax^2 +1)x−x)/(ax^2 +1)) dx =(1/a)∫_0 ^1 dx−(1/a) ∫_0 ^1 ((xdx)/(ax^2 +1)) =(1/a) −(1/(2a^2 )) ∫_0 ^1 ((2ax )/(ax^2 +1)) dx =(1/a) −(1/(2a^2 ))[ln(ax^2 +1)]_0 ^1 =(1/a) −((ln(a+1))/(2a^2 )) ⇒ ((ln(1+a))/a) −(4/(3a)) +((2ln(a+1))/(3a^2 )) ≤ f(a) ≤ ((ln(1+a))/a) ⇒ so we can take v_0 =((ln(1+a))/(2a)) −(2/(3a)) +((ln(1+a))/(3a^2 )) +((ln(1+a))/(2a)) =((ln(1+a))/a) −(2/(3a)) +((ln(1+a))/(3a^2 )) ⇒f(a) ∼ ((ln(1+a))/a) −(2/(3a)) +((ln(1+a))/(3a^2 )) with error δ = (1/2){ (4/(3a))−((2ln(a+1))/(3a^2 ))}$
$1) w e h a v e x - \frac{x^{3}}{6} ⩽ s i n x ⩽ x \Rightarrow 2 x - \frac{8 x^{3}}{6} ⩽ s i n (2 x) ⩽ 2 x \Rightarrow$ $2 x - \frac{4}{3} x^{3} ⩽ s i n (2 x) ⩽ 2 x \Rightarrow \frac{2 x - \frac{4}{3} x^{3}}{1 + {a x}^{2}} ⩽ \frac{s i n (2 x)}{1 + {a x}^{2}} ⩽ \frac{2 x}{1 + {a x}^{2}} \Rightarrow$ $\int_{0}^{1} \frac{2 x d x}{1 + {a x}^{2}} - \frac{4}{3} \int_{0}^{1} \frac{x^{3}}{1 + {a x}^{2}} d x ⩽ \int_{0}^{1} \frac{s i n (2 x)}{1 + {a x}^{2}} d x ⩽ \int_{0}^{1} \frac{2 x}{1 + {a x}^{2}} d x$ $\int_{0}^{1} \frac{2 x d x}{{a x}^{2} + 1} = \frac{1}{a} \int_{0}^{1} \frac{2 a x d x}{{a x}^{2} + 1} = \frac{1}{a} {[l n ({a x}^{2} + 1)]}_{0}^{1} = \frac{l n (1 + a)}{a} (w e s u p p o s e a \neq 0)$ $\int_{0}^{1} \frac{x^{3}}{{a x}^{2} + 1} d x = \frac{1}{a} \int_{0}^{1} \frac{({a x}^{2} + 1) x - x}{{a x}^{2} + 1} d x = \frac{1}{a} \int_{0}^{1} d x - \frac{1}{a} \int_{0}^{1} \frac{x d x}{{a x}^{2} + 1}$ $= \frac{1}{a} - \frac{1}{2 a^{2}} \int_{0}^{1} \frac{2 a x}{{a x}^{2} + 1} d x = \frac{1}{a} - \frac{1}{2 a^{2}} {[l n ({a x}^{2} + 1)]}_{0}^{1} = \frac{1}{a} - \frac{l n (a + 1)}{2 a^{2}} \Rightarrow$ $\frac{l n (1 + a)}{a} - \frac{4}{3 a} + \frac{2 l n (a + 1)}{3 a^{2}} ⩽ f (a) ⩽ \frac{l n (1 + a)}{a} \Rightarrow$ $s o w e c a n t a k e v_{0} = \frac{l n (1 + a)}{2 a} - \frac{2}{3 a} + \frac{l n (1 + a)}{3 a^{2}} + \frac{l n (1 + a)}{2 a}$ $= \frac{l n (1 + a)}{a} - \frac{2}{3 a} + \frac{l n (1 + a)}{3 a^{2}} \Rightarrow f (a) \sim \frac{l n (1 + a)}{a} - \frac{2}{3 a} + \frac{l n (1 + a)}{3 a^{2}}$ $w i t h e r r o r δ = \frac{1}{2} {\frac{4}{3 a} - \frac{2 l n (a + 1)}{3 a^{2}}}$
Commented bymaxmathsup by imad last updated on 02/Jun/19

$2) \int_{0}^{1} \frac{s i n (2 x)}{1 + 2 x^{2}} d x = f (2) a n d f (2) \sim \frac{l n (3)}{2} - \frac{1}{3} + \frac{l n (3)}{12}$ $= \frac{6 l n (3) - 4 + l n (3)}{12} = \frac{7 l n (3) - 4}{12} \Rightarrow \int_{0}^{1} \frac{s i n (2 x)}{1 + 2 x^{2}} d x \sim \frac{7}{12} l n (3) - \frac{1}{3}$
Commented bymaxmathsup by imad last updated on 02/Jun/19

$\int_{0}^{1} \frac{s i n (2 x)}{1 + 2 x^{2}} d x \sim 0, 3$
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