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Question Number 80846 by TawaTawa last updated on 07/Feb/20

Commented by mathmax by abdo last updated on 07/Feb/20
$we havw ∫_(−π) ^π costdt =2∫_0 ^π cost dt =2[sint]_0 ^π =0 ∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ =(1/2) ∫_0 ^(π/2) ((cosθ)/(1−cosθ sinθ))dθ=_(tan((θ/2))=t) =(1/2) ∫_0 ^1 (((1−t^2 )/(1+t^2 ))/(1−((1−t^2 )/(1+t^2 ))×((2t)/(1+t^2 ))))×((2dt)/(1+t^2 )) =∫_0 ^1 ((1−t^2 )/((1+t^2 )^2 {1−((2t(1−t^2 ))/((1+t^2 )^2 ))}))dt =∫_0 ^1 ((1−t^2 )/((1+t^2 )^2 −2t+2t^3 ))dt =∫_0 ^1 ((1−t^2 )/(t^4 +2t^2 +1−2t +2t^3 ))dt let decompose F(t)=((1−t^2 )/(t^4 +2t^3 +2t^2 −2t +1)) the roots of p(x)=t^4 +2t^3 +2t^2 −2t +1 are α(1+i) ,α(1−i) ,(α+1)(−1+i) and (α+1)(−1−i) with α=0,366 ⇒F(t)=((1−t^2 )/((t−α(1+i)(t−α(1−i))(t−(α+1)(−1+i))(t−(α+1)(−1−i)})) =((1−t^2 )/({t^2 −2Re(α(1+i))t +∣α(1+i)∣^2 }{t^2 −2Re(α+1)(−1+i)t +∣(α+1)(−1+i)∣^2 )) =((1−t^2 )/({t^2 −2αt +2α^2 ){t^2 +2 (α+1)t +2(α+1)^2 })) =((at +b)/(t^2 −2αt +2α^2 )) +((ct +d)/(t^2 +2(α+1)t +2(α+1)^2 )) rest to find the coefficient a ,b... any way we have I=[ln(x+(√(1+x^2 )))]_0 ^(∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ) =ln(∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ +(√(1+(∫_0 ^(π/2) ((cosθ)/(2−sin(2θ)))dθ)^2 )))$
$w e h a v w \int_{- π}^{π} c o s t d t = 2 \int_{0}^{π} c o s t d t = 2 {[s i n t]}_{0}^{π} = 0$ $\int_{0}^{\frac{π}{2}} \frac{c o s θ}{2 - s i n (2 θ)} d θ = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{c o s θ}{1 - c o s θ s i n θ} d θ =_{t a n (\frac{θ}{2}) = t}$ $= \frac{1}{2} \int_{0}^{1} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{1 - \frac{1 - t^{2}}{1 + t^{2}} \times \frac{2 t}{1 + t^{2}}} \times \frac{2 d t}{1 + t^{2}} = \int_{0}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{2} {1 - \frac{2 t (1 - t^{2})}{{(1 + t^{2})}^{2}}}} d t$ $= \int_{0}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{2} - 2 t + 2 t^{3}} d t = \int_{0}^{1} \frac{1 - t^{2}}{t^{4} + 2 t^{2} + 1 - 2 t + 2 t^{3}} d t$ $l e t d e c o m p o s e F (t) = \frac{1 - t^{2}}{t^{4} + 2 t^{3} + 2 t^{2} - 2 t + 1}$ $t h e r o o t s o f p (x) = t^{4} + 2 t^{3} + 2 t^{2} - 2 t + 1 a r e$ $α (1 + i), α (1 - i), (α + 1) (- 1 + i) a n d (α + 1) (- 1 - i)$ $w i t h α = 0, 366 \Rightarrow F (t) = \frac{1 - t^{2}}{(t - α (1 + i) (t - α (1 - i)) (t - (α + 1) (- 1 + i)) (t - (α + 1) (- 1 - i)}}$ $= \frac{1 - t^{2}}{{t^{2} - 2 R e (α (1 + i)) t + ∣ α (1 + i) ∣^{2}} {t^{2} - 2 R e (α + 1) (- 1 + i) t + ∣ (α + 1) (- 1 + i) ∣^{2}}$ $= \frac{1 - t^{2}}{{t^{2} - 2 α t + 2 α^{2}) {t^{2} + 2 (α + 1) t + 2 {(α + 1)}^{2}}}$ $= \frac{a t + b}{t^{2} - 2 α t + 2 α^{2}} + \frac{c t + d}{t^{2} + 2 (α + 1) t + 2 {(α + 1)}^{2}}$ $r e s t t o f i n d t h e c o e f f i c i e n t a, b . . . a n y w a y w e h a v e$ $I = {[l n (x + \sqrt{1 + x^{2}})]}_{0}^{\int_{0}^{\frac{π}{2}} \frac{c o s θ}{2 - s i n (2 θ)} d θ}$ $= l n (\int_{0}^{\frac{π}{2}} \frac{c o s θ}{2 - s i n (2 θ)} d θ + \sqrt{1 + {(\int_{0}^{\frac{π}{2}} \frac{c o s θ}{2 - s i n (2 θ)} d θ)}^{2}})$
Commented by TawaTawa last updated on 07/Feb/20

$God bless you sir$
Commented by mathmax by abdo last updated on 08/Feb/20

$y o u a r e w e l c o m e m i s s t a w a w h a t s y o u r r e a l n a m e . . .$
Commented by TawaTawa last updated on 08/Feb/20

$Tawa$
	Answered by Kamel Kamel last updated on 08/Feb/20

	Commented by TawaTawa last updated on 08/Feb/20

	$God bless you sir$
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