let f(x)=∫_0 ^(π/2) (dt/(1+xsint)) with x>−1 1) calculate f(o) ,f(1) and f(2) 2) give f at form of function


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Question Number 51997 by maxmathsup by imad last updated on 01/Jan/19

$l e t f (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x s i n t} w i t h x > - 1$ $1) c a l c u l a t e f (o), f (1) a n d f (2)$ $2) g i v e f a t f o r m o f f u n c t i o n$
Commented by maxmathsup by imad last updated on 02/Jan/19
$1) we have f(0)=∫_0 ^(π/2) dt =(π/2) f(1) =∫_0 ^(π/2) (dt/(1+sint)) =_(tan((t/2))=u) ∫_0 ^1 (1/(1+((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =2 ∫_0 ^1 (du/(1+u^2 +2u)) =2∫_0 ^1 (du/((u+1)^2 )) =[−(2/(u+1))]_0 ^1 =−2((1/2) −1) =−1+2 =1 ⇒f(1)=1 f(2) =∫_0 ^(π/2) (dt/(1+2sint)) =_(tan((t/2))=u) ∫_0 ^1 (1/(1+2((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =2 ∫_0 ^1 (du/(1+u^2 +4u)) =2 ∫_0 ^1 (du/(u^2 +4u +1)) .roots of u^2 +4u +1 Δ^′ =2^2 −1 =3 ⇒u_1 =−2+(√3) and u_2 =−2−(√3) f(2) =2 ∫_0 ^1 (du/((u−u_1 )(u−u_2 ))) =(2/(u_1 −u_2 )) ∫_0 ^1 ((1/(u−u_1 )) −(1/(u−u_2 )))du =(2/(2(√3))) ∫_0 ^1 { (1/(u−u_1 )) −(1/(u−u_2 ))}du =(1/(√3))[ln∣((u−u_1 )/(u−u_2 ))∣]_0 ^1 =(1/(√3)){ln∣((1−u_1 )/(1−u_2 ))∣−ln∣(u_1 /u_2 )∣} =(1/(√3)){ln∣((3−(√3))/(3+(√3)))∣−ln∣((2−(√3))/(2+(√3)))∣ .$
$1) w e h a v e f (0) = \int_{0}^{\frac{π}{2}} d t = \frac{π}{2}$ $f (1) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + s i n t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{1} \frac{1}{1 + \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = 2 \int_{0}^{1} \frac{d u}{1 + u^{2} + 2 u}$ $= 2 \int_{0}^{1} \frac{d u}{{(u + 1)}^{2}} = {[- \frac{2}{u + 1}]}_{0}^{1} = - 2 (\frac{1}{2} - 1) = - 1 + 2 = 1 \Rightarrow f (1) = 1$ $f (2) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + 2 s i n t} =_{t a n (\frac{t}{2}) = u} \int_{0}^{1} \frac{1}{1 + 2 \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= 2 \int_{0}^{1} \frac{d u}{1 + u^{2} + 4 u} = 2 \int_{0}^{1} \frac{d u}{u^{2} + 4 u + 1} . r o o t s o f u^{2} + 4 u + 1$ $Δ^{^{'}} = 2^{2} - 1 = 3 \Rightarrow u_{1} = - 2 + \sqrt{3} a n d u_{2} = - 2 - \sqrt{3}$ $f (2) = 2 \int_{0}^{1} \frac{d u}{(u - u_{1}) (u - u_{2})} = \frac{2}{u_{1} - u_{2}} \int_{0}^{1} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u$ $= \frac{2}{2 \sqrt{3}} \int_{0}^{1} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} d u = \frac{1}{\sqrt{3}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{0}^{1} = \frac{1}{\sqrt{3}} {l n ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣ - l n ∣ \frac{u_{1}}{u_{2}} ∣}$ $= \frac{1}{\sqrt{3}} {l n ∣ \frac{3 - \sqrt{3}}{3 + \sqrt{3}} ∣ - l n ∣ \frac{2 - \sqrt{3}}{2 + \sqrt{3}} ∣ .$
Commented by maxmathsup by imad last updated on 02/Jan/19
$2)changement tan((t/2))=u give f(x)=∫_0 ^1 (1/(1+x((2u)/(1+u^2 )))) ((2du)/(1+u^2 )) =2∫_0 ^1 (du/(1+u^2 +2xu)) =2 ∫_0 ^1 (du/(u^2 +2xu +1)) let p(u)=u^2 +2xu +1 Δ^′ =x^2 −1 case1 Δ^′ >0 ⇔∣x∣>1 ⇒u_1 =−x+(√(x^2 −1)) and u_2 =−x−(√(x^2 −1)) ⇒f(x)=2 ∫_0 ^1 (du/((u−u_1 )(u−u_2 ))) =(2/(u_1 −u_2 ))∫_0 ^1 { (1/(u−u_1 )) −(1/(u−u_2 ))}du =(2/(2(√(1−x^2 )))) [ln∣((u−u_1 )/(u−u_2 ))∣]_0 ^1 =(1/(√(1−x^2 ))){ln∣((1−u_1 )/(1−u_2 ))∣−ln∣(u_1 /u_2 )∣} =(1/(√(1−x^2 ))){ln∣((1+x−(√(x^2 −1)))/(1+x+(√(x^2 −1))))∣−ln∣((x−(√(x^2 −1)))/(x+(√(x^2 −1))))∣} case2 Δ^′ <0 ⇔∣x∣<1 ⇒p(u)=u^2 +2xu +x^2 +1−x^2 =(u+x)^2 +1−x^2 we do the changement u+x=(√(1−x^2 ))α ⇒ f(x) =2 ∫_0 ^1 (du/((u+x)^2 +1−x^2 )) =2 ∫_(x/(√(1−x^2 ))) ^((1+x)/(√(1−x^2 ))) (((√(1−x^2 ))dα)/((1−x^2 )(1+α^2 ))) = (2/(√(1−x^2 ))) [arctan(α)]_(x/(√(1−x^2 ))) ^((1+x)/(√(1−x^2 ))) =(2/(√(1−x^2 ))) { arctan((√((1+x)/(1−x))))−arctan((x/(√(1−x^2 ))))} .$
$2) c h a n g e m e n t t a n (\frac{t}{2}) = u g i v e f (x) = \int_{0}^{1} \frac{1}{1 + x \frac{2 u}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= 2 \int_{0}^{1} \frac{d u}{1 + u^{2} + 2 x u} = 2 \int_{0}^{1} \frac{d u}{u^{2} + 2 x u + 1} l e t p (u) = u^{2} + 2 x u + 1$ $Δ^{^{'}} = x^{2} - 1$ $c a s e 1 Δ^{^{'}} > 0 \Leftrightarrow∣ x ∣> 1 \Rightarrow u_{1} = - x + \sqrt{x^{2} - 1} a n d u_{2} = - x - \sqrt{x^{2} - 1}$ $\Rightarrow f (x) = 2 \int_{0}^{1} \frac{d u}{(u - u_{1}) (u - u_{2})} = \frac{2}{u_{1} - u_{2}} \int_{0}^{1} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} d u$ $= \frac{2}{2 \sqrt{1 - x^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{0}^{1} = \frac{1}{\sqrt{1 - x^{2}}} {l n ∣ \frac{1 - u_{1}}{1 - u_{2}} ∣ - l n ∣ \frac{u_{1}}{u_{2}} ∣}$ $= \frac{1}{\sqrt{1 - x^{2}}} {l n ∣ \frac{1 + x - \sqrt{x^{2} - 1}}{1 + x + \sqrt{x^{2} - 1}} ∣ - l n ∣ \frac{x - \sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}} ∣}$ $c a s e 2 Δ^{^{'}} < 0 \Leftrightarrow∣ x ∣< 1 \Rightarrow p (u) = u^{2} + 2 x u + x^{2} + 1 - x^{2} = {(u + x)}^{2} + 1 - x^{2}$ $w e d o t h e c h a n g e m e n t u + x = \sqrt{1 - x^{2}} α \Rightarrow$ $f (x) = 2 \int_{0}^{1} \frac{d u}{{(u + x)}^{2} + 1 - x^{2}} = 2 \int_{\frac{x}{\sqrt{1 - x^{2}}}}^{\frac{1 + x}{\sqrt{1 - x^{2}}}} \frac{\sqrt{1 - x^{2}} d α}{(1 - x^{2}) (1 + α^{2})}$ $= \frac{2}{\sqrt{1 - x^{2}}} {[a r c t a n (α)]}_{\frac{x}{\sqrt{1 - x^{2}}}}^{\frac{1 + x}{\sqrt{1 - x^{2}}}} = \frac{2}{\sqrt{1 - x^{2}}} {a r c t a n (\sqrt{\frac{1 + x}{1 - x}}) - a r c t a n (\frac{x}{\sqrt{1 - x^{2}}})} .$
	Answered by Smail last updated on 02/Jan/19

	$l e t u = t a n (t / 2) \Rightarrow d t = \frac{2 d u}{1 + u^{2}}$ $s i n t = \frac{2 u}{1 + u^{2}}$ $f (x) = 2 \int_{0}^{1} \frac{d u}{(1 + u^{2}) (1 + \frac{2 x u}{1 + u^{2}})}$ $= 2 \int_{0}^{1} \frac{d u}{u^{2} + 2 x u + 1} = 2 \int_{0}^{1} \frac{d u}{{(u + x)}^{2} + 1 - x^{2}}$ $i f - 1 < x ⩽ 1$ $f (x) = \frac{2}{1 - x^{2}} \int_{0}^{1} \frac{d u}{{(\frac{u + x}{\sqrt{1 - x^{2}}})}^{2} + 1}$ $θ = \frac{u + x}{\sqrt{1 - x^{2}}} \Rightarrow d θ = \frac{d u}{\sqrt{1 - x^{2}}}$ $f (x) = \frac{2}{\sqrt{1 - x^{2}}} \int_{x / \sqrt{1 - x^{2}}}^{(1 + x) / \sqrt{1 - x^{2}}} \frac{d θ}{θ^{2} + 1}$ $= \frac{2}{\sqrt{1 - x^{2}}} {[{t a n}^{- 1} (θ)]}_{x / \sqrt{1 - x^{2}}}^{(1 + x) / \sqrt{1 - x^{2}}}$ $= \frac{2}{\sqrt{1 - x^{2}}} ({t a n}^{- 1} (\frac{1 + x}{\sqrt{1 - x^{2}}}) - {t a n}^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}))$
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