find ∫ (√(2+tan^2 θ))dθ


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Question Number 45519 by maxmathsup by imad last updated on 14/Oct/18

$f i n d \int \sqrt{2 + {t a n}^{2} θ} d θ$
Commented by Meritguide1234 last updated on 14/Oct/18

Commented by maxmathsup by imad last updated on 14/Oct/18
$let A =∫ (√(2+tan^2 θ))dθ changement tanθ =(√2)sh(t) give θ =arctan((√2)sh(t)) ⇒dθ =(((√2)ch(t))/(1+2sh^2 t))dt ⇒ A = ∫ (√(2+2sh^2 t))(((√2)ch(t))/(1+2sh^2 t))dt =2 ∫ ((ch^2 t)/(1+2 sh^2 t))dt =2 ∫ (((1+ch(2t))/2)/(1+2 ((ch(2t)−1)/2)))dt = 2 ∫ ((1+ch(2t))/(1+ch(2t)−1))dt =2 ∫ ( (1/(ch(2t))) +1)dt =2t +2 ∫ (dt/((e^(2t) +e^(−2t) )/2)) =2t +4 ∫ (dt/(e^(2t) +e^(−2t) )) but ∫ (dt/(e^(2t) +e^(−2t) )) =_(e^(2t) =u) ∫ (1/(u+u^(−1) )) (du/(2u)) =(1/2) ∫ (du/(u^2 +1)) =(1/2) arctan(u) (1/2)arctan( e^(2t) ) but sh(t)=((tanθ)/(√2)) ⇒t =argsh(((tanθ)/(√2)))=ln(((tanθ)/(√2)) +(√(1+((tan^2 θ)/2)))) 2t =2ln(((tanθ+(√(2+tan^2 θ)))/(√2))) ⇒e^(2t) =(((tanθ+(√(2+tan^2 θ)))/(√2)))^2 ⇒ A =2ln(((tanθ+(√(2+tan^2 θ)))/(√2))) +2 arctan{(((tanθ+(√(2+tan^2 θ)))/(√2)))^2 } +c$
$l e t A = \int \sqrt{2 + {t a n}^{2} θ} d θ c h a n g e m e n t t a n θ = \sqrt{2} s h (t) g i v e$ $θ = a r c t a n (\sqrt{2} s h (t)) \Rightarrow d θ = \frac{\sqrt{2} c h (t)}{1 + 2 {s h}^{2} t} d t \Rightarrow$ $A = \int \sqrt{2 + 2 {s h}^{2} t} \frac{\sqrt{2} c h (t)}{1 + 2 {s h}^{2} t} d t = 2 \int \frac{{c h}^{2} t}{1 + 2 {s h}^{2} t} d t$ $= 2 \int \frac{\frac{1 + c h (2 t)}{2}}{1 + 2 \frac{c h (2 t) - 1}{2}} d t = 2 \int \frac{1 + c h (2 t)}{1 + c h (2 t) - 1} d t$ $= 2 \int (\frac{1}{c h (2 t)} + 1) d t = 2 t + 2 \int \frac{d t}{\frac{e^{2 t} + e^{- 2 t}}{2}} = 2 t + 4 \int \frac{d t}{e^{2 t} + e^{- 2 t}} b u t$ $\int \frac{d t}{e^{2 t} + e^{- 2 t}} =_{e^{2 t} = u} \int \frac{1}{u + u^{- 1}} \frac{d u}{2 u} = \frac{1}{2} \int \frac{d u}{u^{2} + 1} = \frac{1}{2} a r c t a n (u)$ $\frac{1}{2} a r c t a n (e^{2 t}) b u t s h (t) = \frac{t a n θ}{\sqrt{2}} \Rightarrow t = a r g s h (\frac{t a n θ}{\sqrt{2}}) = l n (\frac{t a n θ}{\sqrt{2}} + \sqrt{1 + \frac{{t a n}^{2} θ}{2}})$ $2 t = 2 l n (\frac{t a n θ + \sqrt{2 + {t a n}^{2} θ}}{\sqrt{2}}) \Rightarrow e^{2 t} = {(\frac{t a n θ + \sqrt{2 + {t a n}^{2} θ}}{\sqrt{2}})}^{2} \Rightarrow$ $A = 2 l n (\frac{t a n θ + \sqrt{2 + {t a n}^{2} θ}}{\sqrt{2}}) + 2 a r c t a n {{(\frac{t a n θ + \sqrt{2 + {t a n}^{2} θ}}{\sqrt{2}})}^{2}} + c$
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