calculate ∫_0 ^∞ ((cos(arctanx))/(4+x^2 ))dx


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Question Number 66695 by mathmax by abdo last updated on 18/Aug/19

$c a l c u l a t e \int_{0}^{\infty} \frac{c o s (a r c t a n x)}{4 + x^{2}} d x$
Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19

$L e t n a m e d i t J$ $A s 1 + {t a n}^{2} t = \frac{1}{{c o s}^{2} t} w e h a v e c o s (a r c t a n x) = \frac{1}{\sqrt{1 + x^{2}}}$ $N o w J = \int_{0}^{\infty} \frac{d x}{(4 + x^{2}) \sqrt{1 + x^{2}}}$ $l e t s t a t e x = s h u \Rightarrow d u = \frac{d x}{\sqrt{1 + x^{2}}}$ $J = \int_{0}^{\infty} \frac{d u}{4 + {s h}^{2} u} = \int_{0}^{\infty} \frac{\frac{1}{{c h}^{2} u}}{\frac{4}{{c h}^{2} u} + {t h}^{2} u} d u$ $A s \frac{1}{{c h}^{2} u} = 1 - {t h}^{2} u a n d \frac{d}{d u} (t h u) = \frac{1}{{c h}^{2} u} w e h a v e$ $J = \int_{0}^{\infty} \frac{d (t h u)}{4 - 3 {t h}^{2} u} = \frac{1}{4} . \int_{0}^{\infty} \frac{d (t h u)}{1 - {(\frac{\sqrt{3}}{2} t h u)}^{2}} = \frac{1}{2 \sqrt{3}} . \int_{0}^{\infty} \frac{d (\frac{\sqrt{3}}{2} t h u)}{1 - {(\frac{\sqrt{3}}{2} t h u)}^{2}} = \frac{1}{2 \sqrt{3}} {[a r g t h (\frac{\sqrt{3}}{2} t h u)]}_{0}^{\infty} = \frac{1}{2 \sqrt{3}} a r g t h (\frac{\sqrt{3}}{2})$ $f i n a l l y$ $J = \frac{1}{2 \sqrt{3}} . \frac{1}{2} l n (\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}) = \frac{1}{4 \sqrt{3}} . l n (\frac{4 + 2 \sqrt{3}}{4 - 2 \sqrt{3}}) = \frac{l n (1 + \sqrt{3})}{\sqrt{3}} - \frac{l n 2}{2 \sqrt{3}}$
Commented by mathmax by abdo last updated on 19/Aug/19

$t h a n k s s i r .$
Commented by mathmax by abdo last updated on 19/Aug/19
$let I=∫_0 ^∞ ((cos(arctanx))/(4+x^2 ))dx we know that cos(arctanx)=(1/(√(1+x^2 ))) ⇒ I =∫_0 ^∞ (dx/((x^2 +4)(√(1+x^2 )))) changement x=sht give I =∫_0 ^∞ ((ch(t))/((sh^2 t +4)ch(t)))dt =∫_0 ^∞ (dt/(4+((ch(2t)−1)/2))) =∫_0 ^∞ ((2dt)/(8+ch(2t)−1)) =∫_0 ^∞ ((2dt)/(7+ch(2t))) =∫_0 ^∞ ((2dt)/(7+((e^(2t) +e^(−2t) )/2))) =∫_0 ^∞ ((4dt)/(14+e^(2t) +e^(−2t) )) =_(e^(2t) =u) ∫_1 ^(+∞) (4/(14 +u+u^(−1) )) (du/(2u)) = ∫_1 ^(+∞) ((2du)/(14u +u^2 +1)) =∫_1 ^(+∞) ((2du)/(u^2 +14u +1)) u^2 +14u+1=0→Δ^′ =7^2 −1 =48 ⇒u_1 =−7+(√(48))=−7+4(√3) u_2 =−7−4(√3) ⇒I =∫_1 ^(+∞) ((2du)/((u−u_1 )(u−u_2 ))) =(2/(8(√3)))∫_1 ^(+∞) {(1/(u−u_1 ))−(1/(u−u_2 ))}du =(1/(4(√3)))[ln∣((u−u_1 )/(u−u_2 ))∣]_1 ^(+∞) =(1/(4(√3))){−ln∣((1−u_1 )/(1−u_2 ))∣} =(1/(4(√3))){ln∣((1+7+4(√3))/(1+7−4(√3)))∣} =(1/(4(√3)))ln∣((8+4(√3))/(8−4(√3)))∣ =(1/(4(√3)))ln(((2+(√3))/(2−(√3)))) .$
$l e t I = \int_{0}^{\infty} \frac{c o s (a r c t a n x)}{4 + x^{2}} d x w e k n o w t h a t c o s (a r c t a n x) = \frac{1}{\sqrt{1 + x^{2}}}$ $\Rightarrow I = \int_{0}^{\infty} \frac{d x}{(x^{2} + 4) \sqrt{1 + x^{2}}} c h a n g e m e n t x = s h t g i v e$ $I = \int_{0}^{\infty} \frac{c h (t)}{({s h}^{2} t + 4) c h (t)} d t = \int_{0}^{\infty} \frac{d t}{4 + \frac{c h (2 t) - 1}{2}} = \int_{0}^{\infty} \frac{2 d t}{8 + c h (2 t) - 1}$ $= \int_{0}^{\infty} \frac{2 d t}{7 + c h (2 t)} = \int_{0}^{\infty} \frac{2 d t}{7 + \frac{e^{2 t} + e^{- 2 t}}{2}} = \int_{0}^{\infty} \frac{4 d t}{14 + e^{2 t} + e^{- 2 t}}$ $=_{e^{2 t} = u} \int_{1}^{+ \infty} \frac{4}{14 + u + u^{- 1}} \frac{d u}{2 u} = \int_{1}^{+ \infty} \frac{2 d u}{14 u + u^{2} + 1}$ $= \int_{1}^{+ \infty} \frac{2 d u}{u^{2} + 14 u + 1}$ $u^{2} + 14 u + 1 = 0 \to Δ^{^{'}} = 7^{2} - 1 = 48 \Rightarrow u_{1} = - 7 + \sqrt{48} = - 7 + 4 \sqrt{3}$ $u_{2} = - 7 - 4 \sqrt{3} \Rightarrow I = \int_{1}^{+ \infty} \frac{2 d u}{(u - u_{1}) (u - u_{2})}$ $= \frac{2}{8 \sqrt{3}} \int_{1}^{+ \infty} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} d u = \frac{1}{4 \sqrt{3}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{1}^{+ \infty}$ $= \frac{1}{4 \sqrt{3}} {- l n ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣} = \frac{1}{4 \sqrt{3}} {l n ∣ \frac{1 + 7 + 4 \sqrt{3}}{1 + 7 - 4 \sqrt{3}} ∣} = \frac{1}{4 \sqrt{3}} l n ∣ \frac{8 + 4 \sqrt{3}}{8 - 4 \sqrt{3}} ∣$ $= \frac{1}{4 \sqrt{3}} l n (\frac{2 + \sqrt{3}}{2 - \sqrt{3}}) .$
	Answered by mind is power last updated on 18/Aug/19

	$c o s (a r c t g (x)) = c o s (x) = \frac{1}{\sqrt{1 + {t g}^{2} (x)}} ==> \frac{1}{\sqrt{(1 + x^{2})}} = c o s (a r c t g (x)) . x ⩾ 0$ $= \int_{0}^{+ \infty} \frac{d x}{\sqrt{(1 + x^{2})} (4 + x^{2})}$ $x = s h (t)$ $d x = c h (t)$ $==> \int_{0}^{+ \infty} \frac{c h (t) d t}{c h (t) (4 + {s h}^{2} (t))}$ $w i t h e e^{t} = w =⩾ d t = \frac{1}{w} d w$ $==> \int_{1}^{+ \infty} \frac{d w}{w (4 + \frac{w^{2}}{4} + \frac{1}{4 w^{2}} - 1)} = \int_{- \infty}^{+ \infty} \frac{4 w d w}{(w^{4} + 12 w^{2} + 1)} = \int \frac{4 w}{(w^{2} + 6 - \sqrt{35}) (w^{2} + 6 + \sqrt{35})}$ $\frac{4 w}{(w^{2} + 6 - \sqrt{35)} (w^{2} + 6 + \sqrt{35})} = \frac{2}{\sqrt{35}} (\frac{w}{w^{2} + 6 - \sqrt{35}} - \frac{w}{w^{2} + 6 + \sqrt{35}})$ $==> \int \frac{4 w d w}{w^{4} - 12 w^{2} + 1} = \frac{1}{\sqrt{35}} l n (\frac{w^{2} + 6 - \sqrt{35}}{w^{2} + 6 + \sqrt{35}}) + c$ $==> \int_{1}^{+ \infty} \frac{4 w d w}{w^{4} - 12 w^{2} + 1} = l i \underset{x \to \infty}{m} {[\frac{2}{\sqrt{35}} \ln (\frac{w^{2} + 6 - \sqrt{35}}{w^{2} + 6 + \sqrt{35}})]}_{1}^{x}$ $= {l i m}_{x \to + \infty} \frac{2}{\sqrt{35}} l n (\frac{x^{2} + 6 - \sqrt{35}}{x^{2} + 6 + \sqrt{35}}) + \frac{2}{\sqrt{35}} l n (\frac{7 + \sqrt{35}}{7 - \sqrt{35}}) = \frac{2}{\sqrt{35}} l n (\frac{7 + \sqrt{35}}{7 - \sqrt{35}})$
	Commented by mathmax by abdo last updated on 19/Aug/19

	$t h a n k s s i r .$
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