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Question Number 110056 by mathdave last updated on 27/Aug/20

p r o v e t h a t

\int_{- \infty}^{+ \infty} \frac{\cos^{4} x - {6 \sin}^{2} x \cos^{2} x + \sin^{4} x}{1 + x^{2}} d x = \frac{π}{e^{4}}

Answered by mathmax by abdo last updated on 26/Aug/20

A = \int_{- \infty}^{+ \infty} \frac{\cos^{4} x - \sin^{4} x - {6 \sin}^{2} x \cos^{2} x}{1 + x^{2}} dx \Rightarrow

A = \int_{- \infty}^{+ \infty} \frac{\cos^{2} x - \sin^{2} x - {6 \sin}^{2} x \cos^{2} x}{1 + x^{2}} dx

= \int_{- \infty}^{+ \infty} \frac{\cos (2 x) - 6 {(\frac{1}{2} \sin (2 x))}^{2}}{1 + x^{2}} dx = \int_{- \infty}^{+ \infty} \frac{\cos (2 x) - \frac{3}{2} \sin^{2} (2 x)}{1 + x^{2}} dx

= \int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{2} + 1} dx - \frac{3}{4} \int_{- \infty}^{+ \infty} \frac{1 - \cos (4 x)}{x^{2} + 1} dx

= \int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{2} + 1} ex - \frac{3}{4} (π) + \frac{3}{4} \int_{- \infty}^{+ \infty} \frac{\cos (4 x)}{x^{2} + 1} dx we have

\int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{2} + 1} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{i 2 x}}{x^{2} + 1} dx) and

\int_{- \infty}^{+ \infty} \frac{e^{i 2 z}}{z^{2} + 1} dx = 2 i π \times Res (f, i) = 2 i π \times \frac{e^{- 2}}{2 i} = π e^{- 2} \Rightarrow

\int_{- \infty}^{+ \infty} \frac{\cos (2 x)}{x^{2} + 1} dx = \frac{π}{e^{2}} also \int_{- \infty}^{+ \infty} \frac{\cos (4 x)}{x^{2} + 1} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{i 4 x}}{x^{2} + 1} dx)

and \int_{- \infty}^{+ \infty} \frac{e^{i 4 z}}{z^{2} + 1} dz = 2 i π Re (f, i) = 2 i π \times \frac{e^{- 4}}{2 i} = \frac{π}{e^{4}} \Rightarrow \int_{- \infty}^{+ \infty} \frac{\cos (4 x)}{x^{2} + 1} dx

= \frac{π}{e^{4}} \Rightarrow A = \frac{π}{e^{2}} - \frac{3 π}{4} + \frac{3}{4} . \frac{π}{e^{4}} = \frac{π}{e^{2}} + \frac{3 π}{{4 e}^{4}} - \frac{3 π}{4}

something went wrong in the question \dots!

Commented by mathdave last updated on 27/Aug/20

i w i l l w o r k i t n o w

Commented by mathdave last updated on 26/Aug/20

t h e q u e s t i o n i s v e r y c o r r e c t

Commented by mathmax by abdo last updated on 26/Aug/20

post your answer sir

Commented by mathmax by abdo last updated on 27/Aug/20

sorry i have solved \int_{R} \frac{\cos^{4} x - \sin^{4} x - {6 \sin}^{2} x \cos^{2} x}{1 + x^{2}} dx

not \int_{R} \frac{\cos^{4} x + \sin^{4} x - 6 \sin^{2} x \cos^{2} x}{1 + x^{2}} dx but the method still

the same \dots

Answered by mathdave last updated on 27/Aug/20

s o l u t i o n

l e t

I = \int_{R} \frac{\cos^{4} x + {2 \cos}^{2} x \sin^{2} x + \sin^{4} x - {8 \cos}^{2} x \sin^{2} x}{1 + x^{2}} d x

I = \int_{R} \frac{{(\cos^{2} x + \sin^{2} x)}^{2} - 2 . 2^{2} \sin^{2} x \cos^{2} x}{1 + x^{2}} d x

I = \int_{R} \frac{1 - {2 \sin}^{2} (2 x)}{1 + x^{2}} d x = \int_{- \infty}^{+ \infty} \frac{\cos (4 x)}{1 + x^{2}} d x \dots . . (x)

l e t t = 4 t h e n I (t) = \int_{- \infty}^{+ \infty} \frac{\cos (t x)}{1 + x^{2}} d x \dots . . (r)

I (0) = \int_{- \infty}^{+ \infty} \frac{\cos (0)}{1 + x^{2}} d x = \int_{R} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} (x)]}_{- \infty}^{+ \infty}

I (0) = \frac{π}{2} - (- \frac{π}{2}) = π \dots \dots \dots \dots (1)

d i f f e r e n t i a t e e q u a t i o n (x) u n d e r (t)

I^{'} (t) = - \int_{R} \frac{x \sin (t x)}{1 + x^{2}} d x = \int_{R} \frac{(x^{2} + 1 - 1) \sin (t x)}{x (1 + x^{2})} d x

I^{'} (t) = \int_{- \infty}^{+ \infty} \frac{\sin (t x)}{x} + \int_{R} \frac{\sin (t x)}{x (1 + x^{2})} d x \dots \dots \dots . (2)

l e t I_{1} (t) = \int_{- \infty}^{+ \infty} \frac{\sin (t x)}{x} d x = 2 \int_{0}^{\infty} \frac{\sin (t x)}{x} d x

b u t M a z i d e n t i t y s a y t h a t

I (a) = \int_{0}^{\infty} \frac{\sin (a x)}{x^{n}} d x = \frac{π}{2 Γ (n) \sin (\frac{n π}{2})} a^{n - 1}

h a s n = 1 a n d a = t

∵ I_{1} (t) = 2 \int_{0}^{\infty} \frac{\sin (t x)}{x} d x = 2 ∙ \frac{π t^{(1 - 1)}}{2 Γ (1) \sin (\frac{π}{2})} = 2 ∙ \frac{π}{2} = π \dots . . (x x)

p u t t i n g e q u a t i o n (x x) i n t o e q u a t i o n (2)

I^{'} (t) = - π + \int_{R} \frac{\sin (t x)}{x (1 + x^{2})} d x \dots \dots (3) (a t t = 0)

I^{'} (0) = - π \dots \dots . (4)

a g a i n d i f f e r r e n t i a t e e q u a t i o n (3) u n d e r (t)

I^{″} (t) = \int_{- \infty}^{+ \infty} \frac{\cos (t x)}{1 + x^{2}} d x = I (t) (t h i s i s t h e s a m e a s e q u a t n (r)

t h e n I^{″} (t) - I (t) = 0 (t a k i n g t h e l a p l a c e t r a n s f o r m)

L [I^{″} (t)] - L [(I (t)] = 0

s^{2} I (s) - s I (0) - I^{'} (0) - I (s) = 0

b u t v a l u e s o f I (0) = π a n d I^{'} (0) = - π

s^{2} I (s) - π s + π - I (s) = 0

I (s) [s^{2} + 1] = π [s - 1]

I (s) [(s + 1) (s - 1)] = π [s - 1]

I (s) = \frac{π}{s + 1} (n o w t a k i n g i n v e r s e l a p l a c e t r a n s f o r m)

L^{- 1} [I (s)] = π L^{- 1} [\frac{1}{s + 1}]

I (t) = π e^{- t} (b u t t = 4)

I (4) = π e^{- 4} = \frac{π}{e^{4}}

∵ \int_{- \infty}^{+ \infty} \frac{\cos^{4} x - {6 \cos}^{2} x \sin^{2} x + \sin^{4} x}{1 + x^{2}} d x = \frac{π}{e^{4}} Q . E . D

b y m a t h d a v e (27 / 08 / 2020)

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