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Question Number 79824 by TawaTawa last updated on 28/Jan/20

Commented by TawaTawa last updated on 28/Jan/20

$Thanks for your time sir . God bless you$
Commented by TawaTawa last updated on 28/Jan/20

$Will be waiting sir$
Commented by MJS last updated on 29/Jan/20

$sorry but changment t = \tan \frac{x}{2} gives$ $\int \frac{2 d t}{p (t^{2} + 1) + \sqrt{1 - t^{4}}}$
Commented by mind is power last updated on 29/Jan/20

$t h i s O n e i s n o t e e a s y i w i l l t r y T o o f i n d a / w a y T o g i v e$ $C l o s e d f o r m i f i c a n f i n d i t$
Commented by mathmax by abdo last updated on 29/Jan/20
$let take a try A =∫_0 ^(π/2) (dx/(p+(√(cosx)))) cha7gement tan((x/2))=t give A =∫_0 ^1 (1/(p+(√((1−t^2 )/(1+t^2 )))))×(2/(1+t^2 ))dt =2 ∫_0 ^1 (dt/(p(1+t^2 )+(√(1−t^4 )))) =2 ∫_0 ^1 ((p(1+t^2 )−(√(1−t^4 )))/(p^2 (1+t^2 )^2 −(1−t^4 )))dt =2p ∫_0 ^1 ((t^2 +1)/(p^2 (t^4 +2t^2 +1)−1+t^4 ))dt −2 ∫_0 ^1 ((√(1−t^4 ))/(p^2 (t^4 +2t^2 +1)−1+t^4 ))dt =2p ∫_0 ^1 ((t^2 +1)/((p^2 +1)t^4 +2p^2 t^2 +p^2 −1))dt −2 ∫_0 ^1 ((√(1−t^4 ))/((p^2 +1)t^4 +2p^2 t^2 +p^2 −1))dt =A−B let decompose F(t) =((t^2 +1)/((p^(2 ) +1)t^4 +2p^2 t^2 +p^2 −1)) =((x+1)/((p^2 +1)x^2 +2p^2 x +p^2 −1))=g(x) with x=t^2 Δ^′ =p^4 −(p^4 −1) =1 ⇒x_1 =((−p^2 +1)/(p^2 +1)) =((1−p^2 )/(1+p^2 )) x_2 =((−p^2 −1)/(p^2 +1)) =−1 ⇒g(x)=((x+1)/((p^2 +1)(x+1)(x−((1−p^2 )/(1+p^2 ))))) =(1/((p^2 +1)(x−((1−p^2 )/(1+p^2 ))))) =(1/((p^2 +1)(t^2 −((1−p^2 )/(1+p^2 ))))) if p>1 we get F(t) =(1/((p^2 +1)(t^2 +((p^2 −1)/(p^2 +1))))) we do the changement t=(√((p^2 −1)/(p^2 +1)))u ⇒2p∫_0 ^1 F(t)dt =((2p)/(p^2 +1)) ∫_0 ^(√((p^2 +1)/(p^2 −1))) (1/(u^2 +1))×((p^2 +1)/(p^2 −1))du =((2p)/(p^2 −1)) arctan(√((p^2 +1)/(p^2 −1))) if 0<p<1 we get F(t)=(1/((p^2 +1)(t−a_p )(t+a_p ))) (a_p =(√((1−p^2 )/(1+p^2 ))))=(1/(2a_p (p^2 +1))){(1/(t−a_p ))−(1/(t+a_p ))} ⇒ 2p ∫_0 ^1 F(t)dt =(p/(a_p (p^2 +1))) [ln∣((t−a_p )/(t+a_p ))∣]_0 ^1 =(p/(a_p (p^2 +1)))ln∣((1−a_p )/(1+a_p ))∣ ...be continued...$
$l e t t a k e a t r y A = \int_{0}^{\frac{π}{2}} \frac{d x}{p + \sqrt{c o s x}} c h a 7 g e m e n t t a n (\frac{x}{2}) = t g i v e$ $A = \int_{0}^{1} \frac{1}{p + \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}} \times \frac{2}{1 + t^{2}} d t = 2 \int_{0}^{1} \frac{d t}{p (1 + t^{2}) + \sqrt{1 - t^{4}}}$ $= 2 \int_{0}^{1} \frac{p (1 + t^{2}) - \sqrt{1 - t^{4}}}{p^{2} {(1 + t^{2})}^{2} - (1 - t^{4})} d t = 2 p \int_{0}^{1} \frac{t^{2} + 1}{p^{2} (t^{4} + 2 t^{2} + 1) - 1 + t^{4}} d t$ $- 2 \int_{0}^{1} \frac{\sqrt{1 - t^{4}}}{p^{2} (t^{4} + 2 t^{2} + 1) - 1 + t^{4}} d t$ $= 2 p \int_{0}^{1} \frac{t^{2} + 1}{(p^{2} + 1) t^{4} + 2 p^{2} t^{2} + p^{2} - 1} d t - 2 \int_{0}^{1} \frac{\sqrt{1 - t^{4}}}{(p^{2} + 1) t^{4} + 2 p^{2} t^{2} + p^{2} - 1} d t$ $= A - B l e t d e c o m p o s e F (t) = \frac{t^{2} + 1}{(p^{2} + 1) t^{4} + 2 p^{2} t^{2} + p^{2} - 1}$ $= \frac{x + 1}{(p^{2} + 1) x^{2} + 2 p^{2} x + p^{2} - 1} = g (x) w i t h x = t^{2}$ $Δ^{^{'}} = p^{4} - (p^{4} - 1) = 1 \Rightarrow x_{1} = \frac{- p^{2} + 1}{p^{2} + 1} = \frac{1 - p^{2}}{1 + p^{2}}$ $x_{2} = \frac{- p^{2} - 1}{p^{2} + 1} = - 1 \Rightarrow g (x) = \frac{x + 1}{(p^{2} + 1) (x + 1) (x - \frac{1 - p^{2}}{1 + p^{2}})}$ $= \frac{1}{(p^{2} + 1) (x - \frac{1 - p^{2}}{1 + p^{2}})} = \frac{1}{(p^{2} + 1) (t^{2} - \frac{1 - p^{2}}{1 + p^{2}})}$ $i f p > 1 w e g e t F (t) = \frac{1}{(p^{2} + 1) (t^{2} + \frac{p^{2} - 1}{p^{2} + 1})} w e d o t h e c h a n g e m e n t$ $t = \sqrt{\frac{p^{2} - 1}{p^{2} + 1}} u \Rightarrow 2 p \int_{0}^{1} F (t) d t = \frac{2 p}{p^{2} + 1} \int_{0}^{\sqrt{\frac{p^{2} + 1}{p^{2} - 1}}} \frac{1}{u^{2} + 1} \times \frac{p^{2} + 1}{p^{2} - 1} d u$ $= \frac{2 p}{p^{2} - 1} a r c t a n \sqrt{\frac{p^{2} + 1}{p^{2} - 1}}$ $i f 0 < p < 1 w e g e t F (t) = \frac{1}{(p^{2} + 1) (t - a_{p}) (t + a_{p})}$ $(a_{p} = \sqrt{\frac{1 - p^{2}}{1 + p^{2}}}) = \frac{1}{2 a_{p} (p^{2} + 1)} {\frac{1}{t - a_{p}} - \frac{1}{t + a_{p}}} \Rightarrow$ $2 p \int_{0}^{1} F (t) d t = \frac{p}{a_{p} (p^{2} + 1)} {[l n ∣ \frac{t - a_{p}}{t + a_{p}} ∣]}_{0}^{1} = \frac{p}{a_{p} (p^{2} + 1)} l n ∣ \frac{1 - a_{p}}{1 + a_{p}} ∣$ $. . . b e c o n t i n u e d . . .$
Commented by TawaTawa last updated on 30/Jan/20

$Thanks for your time sir . I appreciate .$
Commented by mathmax by abdo last updated on 30/Jan/20

$y o u a r e w e l c o m e .$
	Answered by Kamel Kamel last updated on 29/Jan/20

	$Abdomathmax dx = \frac{2 dt}{1 + t^{2}} not \frac{2 dt}{\sqrt{1 + t^{2}}} .$ $The closed form of this integral is$ $given with hypergeometric functions .$
	Answered by mind is power last updated on 30/Jan/20

	Commented by TawaTawa last updated on 30/Jan/20

	$Wow, God bless you sir . I appreciate .$
	Commented by mind is power last updated on 30/Jan/20

	$t h a n k y o u, y^{'} r e w e l c o m w i t h e p l e a s u r$
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