If-2f-x-f-x-1-x-sin-x-1-x-Find-1-e-e-f-x-dx-

Question Number 30871 by ajfour last updated on 27/Feb/18

I f 2 f (x) + f (- x) = \frac{1}{x} \sin (x - \frac{1}{x})

F i n d \int_{1 / e}^{e} f (x) d x .

Commented by abdo imad last updated on 27/Feb/18

l e t c o m p l e t e t h e w o r k o f s i r {m r w}_{2}

w i t h f (x) = \frac{1}{3 x} s i n (x - \frac{1}{x}) w e h a v e

\int_{\frac{1}{e}}^{e} f (x) d x = \frac{1}{3} \int_{\frac{1}{e}}^{e} \frac{1}{x} s i n (x - \frac{1}{x}) d x l e t u s e t h e c h . x = \frac{1}{t} \Rightarrow

3 \int_{\frac{1}{e}}^{e} f (x) d x = - \int_{\frac{1}{e}}^{e} t s i n (\frac{1}{t} - t) \frac{- d t}{t^{2}} = \int_{\frac{1}{e}}^{e} t s i n (\frac{1}{t} - t) d t

= - \int_{\frac{1}{e}}^{e} t s i n (t - \frac{1}{t}) d t = - 3 \int_{\frac{1}{e}}^{e} f (x) d x \Rightarrow

6 \int_{\frac{1}{e}}^{e} f (x) d x \Rightarrow \int_{\frac{1}{e}}^{e} f (x) d x = 0 (i f w e w o r k i n a c o r p s

w i t h c a r e c t e r i s t i c \neq 6)

Commented by ajfour last updated on 28/Feb/18

T h a n k y o u S i r .

Answered by mrW2 last updated on 27/Feb/18

2 f (x) + f (- x) = \frac{1}{x} \sin (x - \frac{1}{x}) \dots (i)

\Rightarrow 2 f (- x) + f (x) = - \frac{1}{x} \sin (- x + \frac{1}{x}) \dots (i i)

2 (i) - (i i) :

3 f (x) = \frac{2}{x} \sin (x - \frac{1}{x}) - \frac{1}{x} \sin (x - \frac{1}{x})

\Rightarrow f (x) = \frac{1}{3 x} \sin (x - \frac{1}{x})

\int_{1 / e}^{e} f (x) d x = \frac{1}{3} \int \frac{1}{x} \sin (x - \frac{1}{x}) d x

l e t u = x - \frac{1}{x}

u^{2} = x^{2} + \frac{1}{x^{2}} - 2

u^{2} + 4 = x^{2} + \frac{1}{x^{2}} + 2 = {(x + \frac{1}{x})}^{2}

\sqrt{u^{2} + 2^{2}} = x + \frac{1}{x}

d u = (1 + \frac{1}{x^{2}}) d x = (x + \frac{1}{x}) \frac{d x}{x} = \sqrt{u^{2} + 2^{2}} \frac{d x}{x}

\frac{d x}{x} = \frac{d u}{\sqrt{u^{2} + 2^{2}}}

\int_{1 / e}^{e} f (x) d x = \frac{1}{3} \int \frac{1}{x} \sin (x - \frac{1}{x}) d x

= \frac{1}{3} \int_{\frac{1}{e} - e}^{e - \frac{1}{e}} \frac{\sin u}{\sqrt{u^{2} + 2^{2}}} d u

= \frac{1}{3} \int_{\frac{1}{e} - e}^{0} \frac{\sin u}{\sqrt{u^{2} + 2^{2}}} d u + \frac{1}{3} \int_{0}^{e - \frac{1}{e}} \frac{\sin u}{\sqrt{u^{2} + 2^{2}}} d u

= - \frac{1}{3} \int_{0}^{e - \frac{1}{e}} \frac{\sin u}{\sqrt{u^{2} + 2^{2}}} d u + \frac{1}{3} \int_{0}^{e - \frac{1}{e}} \frac{\sin u}{\sqrt{u^{2} + 2^{2}}} d u

= 0

Commented by abdo imad last updated on 27/Feb/18

s i r y o u a r e m o r e n e a r f r o m v a l u e l o o k t h a t

\frac{1}{e} - e = - (e - \frac{1}{e}) a n d t h e f u n c t i o n u \to \frac{s i n u}{\sqrt{u^{2} + 4}} i s o d d s o

\int (\dots) d u = 0 .

Commented by mrW2 last updated on 27/Feb/18

t h a n k y o u s i r!

I {d i d n}^{'} t l o o k a t t h e r a n g e o f v a l u e s .

I w a s c o n c e n t r i a t e d i n h o w t o s o l v e

t h e i n t e g r a l . I t i s h a r d i n d e e d .

Commented by mrW2 last updated on 27/Feb/18

C a n w e s o l v e i t, i f t h e q u e s t i o n i s

f i n d \int f (x) d x ?

Commented by prof Abdo imad last updated on 28/Feb/18

p e r h a p s i w i l l t r y \dots

Commented by ajfour last updated on 28/Feb/18

T h a n k y o u S i r .

Commented by abdo imad last updated on 28/Feb/18

b e c a u s e y o u a r e f a m i l i a r w i t h d a n g e r o u s r o a d s l i k e m e \dots .