let f(a) =∫_(π/4) ^(π/3) (√(a+tan^2 x))dx with a>0 1) find a explicit form of f(a) 2) find also g(a) =∫_(π/4) ^(π/3) (dx/(√(a+tan^2 x))) 3) find the values of ∫_(π/4) ^(π/3) (√(2+tan^2 x))dx and ∫


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Question Number 57665 by maxmathsup by imad last updated on 09/Apr/19

$l e t f (a) = \int_{\frac{π}{4}}^{\frac{π}{3}} \sqrt{a + {t a n}^{2} x} d x w i t h a > 0$ $1) f i n d a e x p l i c i t f o r m o f f (a)$ $2) f i n d a l s o g (a) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d x}{\sqrt{a + {t a n}^{2} x}}$ $3) f i n d t h e v a l u e s o f \int_{\frac{π}{4}}^{\frac{π}{3}} \sqrt{2 + {t a n}^{2} x} d x a n d \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d x}{\sqrt{3 + {t a n}^{2} x}}$
Commented by maxmathsup by imad last updated on 12/Apr/19

$1) c h a n g e m e n t t a n x = \sqrt{a} u g i v e f (a) = \int_{\frac{1}{\sqrt{a}}}^{\frac{\sqrt{3}}{\sqrt{a}}} \sqrt{a} \sqrt{1 + u^{2}} \frac{\sqrt{a}}{1 + {a u}^{2}} d u$ $= a \int_{\frac{1}{\sqrt{a}}}^{\frac{\sqrt{3}}{\sqrt{a}}} \frac{\sqrt{1 + u^{2}}}{1 + {a u}^{2}} d u =_{u = s h (t)} a \int_{a r g s h (\frac{1}{\sqrt{a}})}^{a r g s h (\frac{\sqrt{3}}{\sqrt{a}})} \frac{c h (t) c h t d t}{1 + a {s h}^{2} t}$ $= a \int_{l n (\frac{1}{\sqrt{a}} + \sqrt{1 + \frac{1}{a}})}^{l n (\frac{\sqrt{3}}{\sqrt{a}} + \sqrt{1 + \frac{3}{a}})} \frac{\frac{1 + c h (2 t)}{2}}{1 + a \frac{c h (2 t) - 1}{2}} d t = a \int_{α}^{β} \frac{1 + c h (2 t)}{2 - a + a c h (2 t)} d t$ $= a \int_{α}^{β} \frac{1 + \frac{e^{2 t} + e^{- 2 t}}{2}}{2 - a + a \frac{e^{2 t} + e^{- 2 t}}{2}} d t = a \int_{α}^{β} \frac{2 + e^{2 t} + e^{- 2 t}}{4 - 2 a + a e^{2 t} + a e^{- 2 t}} d t$ $=_{e^{2 t} = u} a \int_{e^{2 κ}}^{e^{2 β}} \frac{2 + u + u^{- 1}}{4 - 2 a + a u + {a u}^{- 1}} \frac{d u}{2 u}$ $= a \int_{{(\frac{1}{\sqrt{a}} + \sqrt{1 + \frac{1}{a}})}^{2}}^{{(\frac{\sqrt{3}}{\sqrt{a}} + \sqrt{1 + \frac{3}{a}})}^{2}} \frac{2 u + u^{2} + 1}{2 u^{2} (4 - 2 a) + 2 {a u}^{3} + 2 a u} d u$ $= \frac{a}{2} \int_{{(. . .)}^{2}}^{{(. . .)}^{2}} \frac{u^{2} + 2 u + 1}{u ({a u}^{2} + (4 - 2 a) u + a)} d u l e t d e c o m p o s e$ $F (u) = \frac{u^{2} + 2 u + 1}{u ({a u}^{2} + (4 - 2 a) u + a)}$ $r o o t s o f p (u) = {a u}^{2} + (4 - 2 a) u + a$ $Δ^{^{'}} = {(2 - a)}^{2} - a^{2} = a^{2} - 4 a + 4 - a^{2} = 4 (1 - a)$ $c a s e 1 0 < a < 1 \Rightarrow u_{1} = \frac{- 2 + a + 2 \sqrt{1 - a}}{a} a n d u_{2} = \frac{- 2 + a - 2 \sqrt{1 - a}}{a}$ $F (u) = \frac{u^{2} + 2 u + 1}{a u (u - u_{1}) (u - u_{2})} = \frac{λ_{0}}{u} + \frac{λ_{1}}{u - u_{1}} + \frac{λ_{2}}{u - u_{2}}$ $λ_{0} = {l i m}_{u \to 0} u F (u) = \frac{1}{a}$ $λ_{1} = {l i m}_{u \to u_{1}} (u - u_{1}) F (u) = \frac{u_{1}^{2} + 2 u_{1} + 1}{{a u}_{1} 4 \frac{\sqrt{1 - a}}{a}} = \frac{u_{1}^{2} + 2 u_{1} + 1}{4 u_{1} \sqrt{1 - a}}$ $λ_{2} = {l i m}_{u \to u_{2}} (u - u_{2}) F (u) = \frac{u_{2}^{2} + 2 u_{2} + 1}{{a u}_{2} (- 4) \frac{\sqrt{1 - a}}{a}} = - \frac{u_{2}^{2} + 2 u_{2} + 1}{u_{2} \sqrt{1 - a}}$ $r e s t t o e n d c a l c u l u s \Rightarrow \int F (u) d u = λ_{0} l n ∣ u ∣ + λ_{1} l n ∣ u - u_{1} ∣ + λ_{2} l n ∣ u - u_{2} ∣ + c \Rightarrow$
Commented by maxmathsup by imad last updated on 12/Apr/19

$\Rightarrow f (a) = \frac{a}{2} {[λ_{0} l n ∣ u ∣ + λ_{1} l n ∣ u - u_{1} ∣ + λ_{2} l n ∣ u - u_{2} ∣]}_{{(\frac{1 + \sqrt{1 + a}}{\sqrt{a}})}^{2}}^{{(\frac{\sqrt{3} + \sqrt{3 + a}}{\sqrt{a}})}^{2}}$ $= a λ_{0} l n (\frac{\sqrt{3} + \sqrt{3 + a}}{\sqrt{a}}) + \frac{a λ_{1}}{2} l n ∣ {(\frac{\sqrt{3} + \sqrt{3 + a}}{\sqrt{a}})}^{2} - u_{1} ∣ + \frac{a λ_{2}}{2} l n ∣ {(\frac{\sqrt{3} + \sqrt{3 + a}}{\sqrt{a}})}^{2} - u_{2} ∣$ $- a λ_{0} l n (\frac{1 + \sqrt{1 + a}}{\sqrt{a}}) - \frac{a λ_{1}}{2} l n ∣ {(\frac{1 + \sqrt{1 + a}}{\sqrt{a}})}^{2} - u_{1} ∣ - \frac{a λ_{2}}{2} l n ∣ {(\frac{1 + \sqrt{1 + a}}{\sqrt{a}})}^{2} - u_{2} ∣ .$ $λ_{i} a n d u_{i} a r e k n o w n .$
Commented by maxmathsup by imad last updated on 12/Apr/19

$2) r e s t t o f i n d f (a) i f a > 1 . . . .$ $w e h a v e f^{^{'}} (a) = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d x}{2 \sqrt{a + {t a n}^{2} x}} = \frac{1}{2} g (a) \Rightarrow g (a) = 2 f^{^{'}} (a) . . . .$
Commented by maxmathsup by imad last updated on 12/Apr/19
$3) let begin from the begining we have A =∫_(π/4) ^(π/3) (√(2+tan^2 x))dx changement tanx =(√2)u give A =∫_(1/(√2)) ^((√3)/(√2)) (√2)(√(1+u^2 )) ((√2)/(1+2u^2 )) du =2 ∫_(1/(√2)) ^((√3)/(√2)) ((√(1+u^2 ))/(1+2u^2 )) du =_(u =sh(t)) 2 ∫_(argsh((1/(√2)))) ^(argsh(((√3)/(√2)))) ((ch(t) ch(t)dt)/(1+2sh^2 (t))) =2 ∫_(ln((1/(√2)) +(√(1+(1/2))))) ^(ln(((√3)/(√2)) +(√(1+(3/2))))) (((1+ch(2t))/2)/(1+2 ((ch(2t)−1)/2))) dt = 2 ∫_(ln(((1+(√3))/(√2)))) ^(ln((((√(3 )) +(√5))/(√2)))) ((1+ch(2t))/(2+2ch(2t)−2))dt =2 ∫_α ^β ((1+((e^(2t) +e^(−2t) )/2))/((e^(2t) +e^(−2t) )/2)) dt =2 ∫_α ^β ((2 +e^(2t) +e^(−2t) )/(e^(2t) +e^(−2t) )) dt =_(e^(2t) =u) 2 ∫_e^(2α) ^e^(2β) ((2 +u +u^(−1) )/(u +u^(−1) )) (du/(2u)) = ∫_((((1+(√3))/(√2)))^2 ) ^(((((√3) +(√5))/(√2)))^2 ) ((2u+u^2 +1)/(u(u^2 +1))) du let decompose F(u) =((u^2 +2u +1)/(u(u^2 +1))) F(u) = (a/u) +((bu +c)/(u^2 +1)) a =lim_(u→0) uF(u) =1 lim_(u→+∞) u F(u) =1 =a+b ⇒b =0 ⇒F(u) =(1/u) +(c/(u^2 +1)) F(1) =2 = 1+(c/2) ⇒(c/2) =1 ⇒c =2 ⇒ F(u) =(1/u) +(2/(1+u^2 )) ⇒ ∫ F(u)du =ln∣u∣ +2arctan(u)+c ⇒ ∫_((((1+(√3))/(√2)))^2 ) ^(((((√3) +(√5))/(√2)))^2 ) ((u^2 +2u +1)/(u(u^2 +1))) du =[ln∣u∣ +2arctan(u)]_((((1+(√3))/(√2)))^2 ) ^(((((√3) +(√5))/(√2)))^2 ) ⇒ A=2ln((((√3) +(√5))/(√2))) +2arctan{((((√3) +(√5))/(√2)))^2 }−2ln(((1+(√3))/(√2)))−2 arctan{(((1+(√3))/(√2)))^2 } .$
$3) l e t b e g i n f r o m t h e b e g i n i n g w e h a v e A = \int_{\frac{π}{4}}^{\frac{π}{3}} \sqrt{2 + {t a n}^{2} x} d x$ $c h a n g e m e n t t a n x = \sqrt{2} u g i v e A = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{\sqrt{2}}} \sqrt{2} \sqrt{1 + u^{2}} \frac{\sqrt{2}}{1 + 2 u^{2}} d u$ $= 2 \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{\sqrt{2}}} \frac{\sqrt{1 + u^{2}}}{1 + 2 u^{2}} d u =_{u = s h (t)} 2 \int_{a r g s h (\frac{1}{\sqrt{2}})}^{a r g s h (\frac{\sqrt{3}}{\sqrt{2}})} \frac{c h (t) c h (t) d t}{1 + 2 {s h}^{2} (t)}$ $= 2 \int_{l n (\frac{1}{\sqrt{2}} + \sqrt{1 + \frac{1}{2}})}^{l n (\frac{\sqrt{3}}{\sqrt{2}} + \sqrt{1 + \frac{3}{2}})} \frac{\frac{1 + c h (2 t)}{2}}{1 + 2 \frac{c h (2 t) - 1}{2}} d t = 2 \int_{l n (\frac{1 + \sqrt{3}}{\sqrt{2}})}^{l n (\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})} \frac{1 + c h (2 t)}{2 + 2 c h (2 t) - 2} d t$ $= 2 \int_{α}^{β} \frac{1 + \frac{e^{2 t} + e^{- 2 t}}{2}}{\frac{e^{2 t} + e^{- 2 t}}{2}} d t = 2 \int_{α}^{β} \frac{2 + e^{2 t} + e^{- 2 t}}{e^{2 t} + e^{- 2 t}} d t =_{e^{2 t} = u} 2 \int_{e^{2 α}}^{e^{2 β}} \frac{2 + u + u^{- 1}}{u + u^{- 1}} \frac{d u}{2 u}$ $= \int_{{(\frac{1 + \sqrt{3}}{\sqrt{2}})}^{2}}^{{(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{2}} \frac{2 u + u^{2} + 1}{u (u^{2} + 1)} d u l e t d e c o m p o s e F (u) = \frac{u^{2} + 2 u + 1}{u (u^{2} + 1)}$ $F (u) = \frac{a}{u} + \frac{b u + c}{u^{2} + 1}$ $a = {l i m}_{u \to 0} u F (u) = 1$ ${l i m}_{u \to + \infty} u F (u) = 1 = a + b \Rightarrow b = 0 \Rightarrow F (u) = \frac{1}{u} + \frac{c}{u^{2} + 1}$ $F (1) = 2 = 1 + \frac{c}{2} \Rightarrow \frac{c}{2} = 1 \Rightarrow c = 2 \Rightarrow F (u) = \frac{1}{u} + \frac{2}{1 + u^{2}} \Rightarrow$ $\int F (u) d u = l n ∣ u ∣ + 2 a r c t a n (u) + c \Rightarrow$ $\int_{{(\frac{1 + \sqrt{3}}{\sqrt{2}})}^{2}}^{{(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{2}} \frac{u^{2} + 2 u + 1}{u (u^{2} + 1)} d u = {[l n ∣ u ∣ + 2 a r c t a n (u)]}_{{(\frac{1 + \sqrt{3}}{\sqrt{2}})}^{2}}^{{(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{2}} \Rightarrow$ $A = 2 l n (\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}) + 2 a r c t a n {{(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{2}} - 2 l n (\frac{1 + \sqrt{3}}{\sqrt{2}}) - 2 a r c t a n {{(\frac{1 + \sqrt{3}}{\sqrt{2}})}^{2}} .$
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