let-f-a-0-1-sin-2x-1-ax-2-dx-with-a-lt-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-s

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 by maxmathsup 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 by maxmathsup by imad last updated on 02/Jun/19

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 by maxmathsup 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 by maxmathsup 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