∫(dx/(a+btan^2 x)) = ?


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Question Number 38074 by ajfour last updated on 21/Jun/18

$\int \frac{d x}{a + b \tan^{2} x} = ?$
Commented by prof Abdo imad last updated on 21/Jun/18
$changement tanx=t give I = ∫ (1/(a+bt^2 )) (dt/(1+t^2 )) =∫ (dt/((1+t^2 )(a+bt^2 ))) let decompose F(t)= (1/((1+t^2 )(a+bt^2 ))) F(t)= ((αt +β)/(t^2 +1)) +((ct +d)/(bt^2 +a)) F(−t)=F(t) ⇒((−αt +β)/(t^2 +1)) +((−ct +d)/(bt^2 +a)) =F(t)⇒ α=c=0 ⇒F(t)=(β/(1+t^2 )) +(d/(bt^2 +a)) lim_(t→+∞) t^2 F(t)=0=β +(d/b) ⇒bβ +d=0 ⇒ d=−bβ ⇒F(t)=(β/(1+t^2 )) −((bβ)/(bt^2 +a)) F(0)= (1/a) =β −((bβ)/a) ⇒1=aβ −bβ ⇒β= (1/(a−b)) if a≠b ⇒F(t)= (1/(a−b)){ (1/(1+t^2 )) −(b/(bt^2 +a))} I = ∫ F(t)dt ⇒(a−b)I= ∫ (dt/(1+t^2 )) −∫ (dt/(t^2 +(a/b))) case1 (a/b)>0 we do the chang.t=(√(a/b)) u ∫ (dt/(t^2 +(a/b))) = ∫ (1/((a/b)(1+u^2 ))) (√(a/b)) du =(b/a) (√(a/b)) arctanu =(√(b/a)) arctan((√(b/a))tanx)⇒ (a−b)I= x −(√(b/a)) arctan((√(b/a))tanx)⇒ I= (1/(a−b)){ x−(√(b/a))arctan((√(b/a))tanx) +λ case2 (a/b)<0 ∫ (dt/(t^2 +(a/b))) =∫ (dt/(t^2 −((√(−(a/b))))^2 )) = (1/(2(√(−(a/b)))))∫ ( (1/(t−(√(−(a/b))))) −(1/(t +(√(−(a/b))))))dt = (1/(2(√(−(a/b)))))ln∣ ((t−(√(−(a/b))))/(t+(√(−(a/b)))))∣ ⇒ I = (1/(a−b)){ x −(1/(2(√(−(a/b))))) ln∣((tan(x) −(√(−(a/b))))/(tanx +(√(−(a/b)))))∣} +λ$
$c h a n g e m e n t t a n x = t g i v e$ $I = \int \frac{1}{a + {b t}^{2}} \frac{d t}{1 + t^{2}} = \int \frac{d t}{(1 + t^{2}) (a + {b t}^{2})}$ $l e t d e c o m p o s e F (t) = \frac{1}{(1 + t^{2}) (a + {b t}^{2})}$ $F (t) = \frac{α t + β}{t^{2} + 1} + \frac{c t + d}{{b t}^{2} + a}$ $F (- t) = F (t) \Rightarrow \frac{- α t + β}{t^{2} + 1} + \frac{- c t + d}{{b t}^{2} + a} = F (t) \Rightarrow$ $α = c = 0 \Rightarrow F (t) = \frac{β}{1 + t^{2}} + \frac{d}{{b t}^{2} + a}$ ${l i m}_{t \to + \infty} t^{2} F (t) = 0 = β + \frac{d}{b} \Rightarrow b β + d = 0 \Rightarrow$ $d = - b β \Rightarrow F (t) = \frac{β}{1 + t^{2}} - \frac{b β}{{b t}^{2} + a}$ $F (0) = \frac{1}{a} = β - \frac{b β}{a} \Rightarrow 1 = a β - b β \Rightarrow β = \frac{1}{a - b}$ $i f a \neq b \Rightarrow F (t) = \frac{1}{a - b} {\frac{1}{1 + t^{2}} - \frac{b}{{b t}^{2} + a}}$ $I = \int F (t) d t \Rightarrow (a - b) I = \int \frac{d t}{1 + t^{2}} - \int \frac{d t}{t^{2} + \frac{a}{b}}$ $c a s e 1 \frac{a}{b} > 0 w e d o t h e c h a n g . t = \sqrt{\frac{a}{b}} u$ $\int \frac{d t}{t^{2} + \frac{a}{b}} = \int \frac{1}{\frac{a}{b} (1 + u^{2})} \sqrt{\frac{a}{b}} d u$ $= \frac{b}{a} \sqrt{\frac{a}{b}} a r c t a n u = \sqrt{\frac{b}{a}} a r c t a n (\sqrt{\frac{b}{a}} t a n x) \Rightarrow$ $(a - b) I = x - \sqrt{\frac{b}{a}} a r c t a n (\sqrt{\frac{b}{a}} t a n x) \Rightarrow$ $I = \frac{1}{a - b} {x - \sqrt{\frac{b}{a}} a r c t a n (\sqrt{\frac{b}{a}} t a n x) + λ$ $c a s e 2 \frac{a}{b} < 0$ $\int \frac{d t}{t^{2} + \frac{a}{b}} = \int \frac{d t}{t^{2} - {(\sqrt{- \frac{a}{b}})}^{2}}$ $= \frac{1}{2 \sqrt{- \frac{a}{b}}} \int (\frac{1}{t - \sqrt{- \frac{a}{b}}} - \frac{1}{t + \sqrt{- \frac{a}{b}}}) d t$ $= \frac{1}{2 \sqrt{- \frac{a}{b}}} l n ∣ \frac{t - \sqrt{- \frac{a}{b}}}{t + \sqrt{- \frac{a}{b}}} ∣ \Rightarrow$ $I = \frac{1}{a - b} {x - \frac{1}{2 \sqrt{- \frac{a}{b}}} l n ∣ \frac{t a n (x) - \sqrt{- \frac{a}{b}}}{t a n x + \sqrt{- \frac{a}{b}}} ∣} + λ$
Commented by prof Abdo imad last updated on 21/Jun/18

$i f a = b I = \frac{1}{a} \int \frac{d x}{1 + {t a n}^{2} x} (a \neq 0)$ $= \frac{1}{a} \int {c o s}^{2} x d x = \frac{1}{a} \int \frac{1 + c o s (2 x)}{2} d x$ $= \frac{x}{2 a} + \frac{1}{4 a} s i n (2 x) + λ .$
Commented by ajfour last updated on 21/Jun/18

$W o w S i r, t h a n k y o u .$
Commented by math khazana by abdo last updated on 21/Jun/18

$n e v e r m i n d s i r A j f o u r .$
	Answered by behi83417@gmail.com last updated on 21/Jun/18

	${t g}^{2} x = \frac{a}{b} t \Rightarrow t g x = \sqrt{\frac{a}{b} t}$ $2 t g x (1 + {t g}^{2} x) d x = \frac{a}{b} d t \Rightarrow$ $2 \sqrt{\frac{a t}{b}} (1 + \frac{a t}{b}) d x = \frac{a}{b} d t \Rightarrow d x = \frac{a d t}{2 \sqrt{\frac{a t}{b}} . (b + a t)}$ $\Rightarrow I = \frac{1}{2 \sqrt{\frac{a}{b}}} \int \frac{\frac{a d t}{\sqrt{t} (b + a t)}}{a + a t} = m \int \frac{d t}{\sqrt{t} (1 + t) (b + a t)}$ $m = {(2 \sqrt{\frac{a}{b}})}^{- 1}, t = u^{2} \Rightarrow d t = 2 u d u$ $I = m \int \frac{2 u d u}{u (1 + u^{2}) (b + {a u}^{2})} = 2 m \int \frac{d u}{(1 + u^{2}) (b + {a u}^{2})} =$ $= 2 m [a . \frac{1}{4 \sqrt{a b}} {t g}^{- 1} (a . \frac{u}{\sqrt{a b}}) - \frac{1}{4} {t g}^{- 1} u] + c o n s t$ $= \frac{1}{2} m [\sqrt{\frac{a}{b}} . {t g}^{- 1} (\sqrt{\frac{a}{b}} u) - {t g}^{- 1} u + c o n s t =$ $= \frac{1}{2} . \frac{1}{2} \sqrt{\frac{b}{a}} . \sqrt{\frac{a}{b}} {t g}^{- 1} (\sqrt{\frac{a}{b}} . \sqrt{\frac{b}{a}} . t g x) -$ $- \frac{1}{4} . \sqrt{\frac{b}{a}} . {t g}^{- 1} (\sqrt{\frac{b}{a}} . t g x) + c o n s t =$ $= \frac{1}{4} x - \frac{1}{4} . \sqrt{\frac{b}{a}} . a r c t g [\sqrt{\frac{b}{a}} . t g x] + c o n s t . ◼$
	Commented by ajfour last updated on 21/Jun/18

	$G o d b l e s s y o u S i r .$
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