1) calculate f(a) =∫_(−∞) ^(+∞) (dx/(x^2 +ax +1)) with ∣a∣<2 2) calculate g(a) =∫_(−∞) ^(+∞) (x/((x^2 +ax+1)^2 )) 3)find values of integrals ∫_(−∞) ^(+∞) (dx/(x^2 +(√2)x +1)) and ∫_(−∞) ^(+∞) (x/((x^2 +(√2)x +1)^2 )) 4) calculate A(θ) = ∫_(−∞) ^(+∞) (dx/(x^2 +2cosθ +1)) θ is a given real.


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Question Number 54376 by Abdo msup. last updated on 02/Feb/19

$1) c a l c u l a t e f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + a x + 1}$ $w i t h ∣ a ∣< 2$ $2) c a l c u l a t e g (a) = \int_{- \infty}^{+ \infty} \frac{x}{{(x^{2} + a x + 1)}^{2}}$ $3) f i n d v a l u e s o f i n t e g r a l s \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + \sqrt{2} x + 1}$ $a n d \int_{- \infty}^{+ \infty} \frac{x}{{(x^{2} + \sqrt{2} x + 1)}^{2}}$ $4) c a l c u l a t e A (θ) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 c o s θ + 1}$ $θ i s a g i v e n r e a l .$
Commented bymaxmathsup by imad last updated on 03/Feb/19

$1) w e h a v e f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + a x + 1} \Rightarrow f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 x \frac{a}{2} + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4}}$ $= \int_{- \infty}^{+ \infty} \frac{d x}{{(x + \frac{a}{2})}^{2} + \frac{4 - a^{2}}{4}} =_{x + \frac{a}{2} = \frac{1}{2} \sqrt{4 - a^{2}} t} \int_{- \infty}^{+ \infty} \frac{1}{\frac{4 - a^{2}}{4} (1 + t^{2})} \frac{\sqrt{4 - a^{2}} d t}{2}$ $= 2 \frac{1}{\sqrt{4 - a^{2}}} \int_{- \infty}^{+ \infty} \frac{d t}{1 + t^{2}} = \frac{2 π}{\sqrt{4 - a^{2}}} \Rightarrow f (a) = \frac{2 π}{\sqrt{4 - a^{2}}} .$ $2) w e h a v e f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} + a x + 1)}^{2}} = - g (a) \Rightarrow$ $g (a) = - f^{^{'}} (a) = - 2 π {({(4 - a^{2})}^{- \frac{1}{2}})}^{^{'}} = - 2 π . (- \frac{1}{2}) (- 2 a) {(4 - a^{2})}^{- \frac{3}{2}}$ $= - 2 π a \frac{1}{(4 - a^{2}) \sqrt{4 - a^{2}}} = \frac{- 2 π a}{(4 - a^{2}) \sqrt{4 - a^{2}}}$
Commented bymaxmathsup by imad last updated on 03/Feb/19

$3) \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + \sqrt{2} x + 1} = f (\sqrt{2}) = \frac{2 π}{\sqrt{4 - 2}} = \frac{2 π}{\sqrt{2}} = π \sqrt{2} .$ $\int_{- \infty}^{+ \infty} \frac{x d x}{{(x^{2} + \sqrt{2} x + 1)}^{2}} = g (\sqrt{2}) = \frac{- 2 π \sqrt{2}}{2 \sqrt{2}} = - π .$
Commented bymaxmathsup by imad last updated on 03/Feb/19

$4) \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2 c o s θ x + 1} = f (2 c o s θ) = \frac{2 π}{\sqrt{4 - 4 {c o s}^{2} θ}} = \frac{2 π}{2 \sqrt{1 - {c o s}^{2} θ}} = \frac{π}{∣ s i n θ ∣}$ $(w e s u p p o s e h e r e θ \neq k π k f r o m Z) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
	$2)(1/2)∫((2x+a−a)/((x^2 +ax+1)^2 ))dx (1/2)∫((d(x^2 +ax+1))/((x^2 +ax+1)^2 ))dx−(a/2)∫(dx/([(x+(a/2))^2 +((√(1−(a^2 /4))) )^2 ]^2 )) (1/2)×((−1)/((x^2 +ax+1)))−(a/2)I_2 I_2 let (x+(a/2))=(√(1−(a^2 /4))) tanθ=ktanθ dx=k sec^2 θdθ so I_2 =∫((ksec^2 θdθ)/([k^2 tan^2 θ+k^2 ]^2 )) ∫((ksec^2 θ)/(k^4 sec^4 θ))dθ (1/k^3 )∫(((1+cos2θ)/2))dθ (1/(2k^3 ))θ+(1/(4k^3 ))sin2θ+c (1/(2k^3 ))tan^(−1) (((x+(a/2))/k))+(1/(4k^3 ))×((2(((x+(a/2))/k)))/(1+(((x+(a/2))/k))^2 ))+c so ((−1)/(2(x^2 +ax+1)))−(a/2)[(1/(2k^3 ))tan^(−1) (((x+(a/2))/k))+(1/(2k^3 ))×(((((x+(a/2))/k)))/(1+(((x+(a/2))/k))^2 ))] ∣((−1)/(2(x^2 +ax+1)))−(a/(4k^3 ))[tan^(−1) (((x+(a/2))/k))+(((x+(a/2))/k)/(1+(((x+(a/2))/k))^2 ))]∣_(−∞) ^∞ =0−(a/(4k^3 ))[{tan^(−1) (∞)−tan^(−1) (−∞)}+0] =−(a/(4k^3 ))×(π/1)=((−aπ)/(4(1−(a^2 /4))^(3/2) )) 3)second part =((−(√2) π)/(4(1−(1/2))^(3/2) ))=((−(√2) π)/4)×(√2) ×(√2) ×(√2) =−π answer$
	$2) \frac{1}{2} \int \frac{2 x + a - a}{{(x^{2} + a x + 1)}^{2}} d x$ $\frac{1}{2} \int \frac{d (x^{2} + a x + 1)}{{(x^{2} + a x + 1)}^{2}} d x - \frac{a}{2} \int \frac{d x}{{[{(x + \frac{a}{2})}^{2} + {(\sqrt{1 - \frac{a^{2}}{4}})}^{2}]}^{2}}$ $\frac{1}{2} \times \frac{- 1}{(x^{2} + a x + 1)} - \frac{a}{2} I_{2}$ $I_{2} l e t (x + \frac{a}{2}) = \sqrt{1 - \frac{a^{2}}{4}} t a n θ = k t a n θ$ $d x = k {s e c}^{2} θ d θ$ $s o I_{2} = \int \frac{{k s e c}^{2} θ d θ}{{[k^{2} {t a n}^{2} θ + k^{2}]}^{2}}$ $\int \frac{{k s e c}^{2} θ}{k^{4} {s e c}^{4} θ} d θ$ $\frac{1}{k^{3}} \int (\frac{1 + c o s 2 θ}{2}) d θ$ $\frac{1}{2 k^{3}} θ + \frac{1}{4 k^{3}} s i n 2 θ + c$ $\frac{1}{2 k^{3}} {t a n}^{- 1} (\frac{x + \frac{a}{2}}{k}) + \frac{1}{4 k^{3}} \times \frac{2 (\frac{x + \frac{a}{2}}{k})}{1 + {(\frac{x + \frac{a}{2}}{k})}^{2}} + c$ $s o$ $\frac{- 1}{2 (x^{2} + a x + 1)} - \frac{a}{2} [\frac{1}{2 k^{3}} {t a n}^{- 1} (\frac{x + \frac{a}{2}}{k}) + \frac{1}{2 k^{3}} \times \frac{(\frac{x + \frac{a}{2}}{k})}{1 + {(\frac{x + \frac{a}{2}}{k})}^{2}}]$ $∣ \frac{- 1}{2 (x^{2} + a x + 1)} - \frac{a}{4 k^{3}} [{t a n}^{- 1} (\frac{x + \frac{a}{2}}{k}) + \frac{\frac{x + \frac{a}{2}}{k}}{1 + {(\frac{x + \frac{a}{2}}{k})}^{2}}] ∣_{- \infty}^{\infty}$ $= 0 - \frac{a}{4 k^{3}} [{{t a n}^{- 1} (\infty) - {t a n}^{- 1} (- \infty)} + 0]$ $= - \frac{a}{4 k^{3}} \times \frac{π}{1} = \frac{- a π}{4 {(1 - \frac{a^{2}}{4})}^{\frac{3}{2}}}$ $3) s e c o n d p a r t = \frac{- \sqrt{2} π}{4 {(1 - \frac{1}{2})}^{\frac{3}{2}}} = \frac{- \sqrt{2} π}{4} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} = - π a n s w e r$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
	$1)∫(dx/(x^2 +2x(a/2)+(a^2 /4)+1−(a^2 /4)))dx ∫(dx/(((√(1−(a^2 /4))) )^2 +(x+(a/2))^2 )) so answer is ∣(1/((√(1−(a^2 /4))) ))tan^(−1) (((x+(a/2))/(√(1−(a^2 /4)))))∣_(−∞) ^∞ [(a^2 /4)<1] =(1/(√(1−(a^2 /4))))×{(π/2)−(−(π/2))}=(π/(√(1−(a^2 /4)))) 3)(π/(√(1−(2/4))))=π(√2)$
	$1) \int \frac{d x}{x^{2} + 2 x \frac{a}{2} + \frac{a^{2}}{4} + 1 - \frac{a^{2}}{4}} d x$ $\int \frac{d x}{{(\sqrt{1 - \frac{a^{2}}{4}})}^{2} + {(x + \frac{a}{2})}^{2}}$ $s o a n s w e r i s$ $∣ \frac{1}{\sqrt{1 - \frac{a^{2}}{4}}} {t a n}^{- 1} (\frac{x + \frac{a}{2}}{\sqrt{1 - \frac{a^{2}}{4}}}) ∣_{- \infty}^{\infty} [\frac{a^{2}}{4} < 1]$ $= \frac{1}{\sqrt{1 - \frac{a^{2}}{4}}} \times {\frac{π}{2} - (- \frac{π}{2})} = \frac{π}{\sqrt{1 - \frac{a^{2}}{4}}}$ $3) \frac{π}{\sqrt{1 - \frac{2}{4}}} = π \sqrt{2}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
	$4)∫_(−∞) ^∞ (dx/(x^2 +b^2 )) [[[b^2 =1+2cosθ] (1/b)∣tan^(−1) ((x/b))∣_(−∞) ^∞ (1/b)[tan^(−1) (∞)−tan^(−1) (−∞)] =(1/(√(1+2cosθ)))×{(π/2)−((−π)/2)} =(π/(√(1+2cosθ)))$
	$4) \int_{- \infty}^{\infty} \frac{d x}{x^{2} + b^{2}} [[[b^{2} = 1 + 2 c o s θ]$ $\frac{1}{b} ∣ {t a n}^{- 1} (\frac{x}{b}) ∣_{- \infty}^{\infty}$ $\frac{1}{b} [{t a n}^{- 1} (\infty) - {t a n}^{- 1} (- \infty)]$ $= \frac{1}{\sqrt{1 + 2 c o s θ}} \times {\frac{π}{2} - \frac{- π}{2}}$ $= \frac{π}{\sqrt{1 + 2 c o s θ}}$
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