∫ ((acos x+b)/((a+bcos x)^2 ))dx = ?


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Question Number 37317 by rahul 19 last updated on 11/Jun/18

$\int \frac{acos x + b}{{(a + b \cos x)}^{2}} d x = ?$
Commented by math khazana by abdo last updated on 12/Jun/18
$changement tan((x/2)) = t give I = ∫ ((a((1−t^2 )/(1+t^2 )) +b)/((a +b((1−t^2 )/(1+t^2 )))^2 )) ((2dt)/(1+t^2 )) =2 ∫ ((a(1−t^2 ) +b +bt^2 )/((1+t^2 )^2 (((a+at^2 +b−bt^2 )^2 )/((1+t^2 )^2 )))) dt = 2 ∫ ((a+b +(b−a)t^2 )/({a+b +(a−b)t^2 }^2 )) dt case 1 a>b I =2 ∫ (((a+b){ 1−(a−b)t^2 })/((a+b)^2 { 1+((a−b)/(a+b))t^2 }^2 )) dt =(2/(a+b)) ∫ ((1−((a−b)/(a+b))t^2 )/({1 +((a−b)/(a+b))t^2 }^2 ))dt changement (√((a−b)/(a+b)))t =x give I = (2/(a+b)) ∫ ((1−x^2 )/({1+x^2 }^2 )) (√((a+b)/(a−b))) dx = (2/(√(a^2 −b^2 ))) ∫ ((1−x^2 )/((1+x^2 )^2 )) dx but ∫ ((1−x^2 )/((1+x^2 )^2 ))dx = ∫ ((1+x^2 −2x^2 )/((1+x^2 )^2 ))dx = ∫ (dx/(1+x^2 )) −∫ ((2x^2 )/((1+x^2 )^2 ))dx =arctanx −∫ ((2x^2 )/((1+x^2 )^2 ))dx by parts u^′ = ((−2x)/((1+x^2 )^2 )) and u =x ∫ ((−2x)/((1+x^2 )^2 )) x dx = (1/(1+x^2 )) x −∫ (1/(1+x^2 )) dx =(x/(1+x^2 )) −arctan(x) ⇒ I = (2/(√(a^2 −b^2 ))){ arctan(x) +(x/(1+x^2 )) −arctan(x)} +c I = ((2x)/((√(a^2 −b^2 ))(1+x^2 )))$
$c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{a \frac{1 - t^{2}}{1 + t^{2}} + b}{{(a + b \frac{1 - t^{2}}{1 + t^{2}})}^{2}} \frac{2 d t}{1 + t^{2}}$ $= 2 \int \frac{a (1 - t^{2}) + b + {b t}^{2}}{{(1 + t^{2})}^{2} \frac{{(a + {a t}^{2} + b - {b t}^{2})}^{2}}{{(1 + t^{2})}^{2}}} d t$ $= 2 \int \frac{a + b + (b - a) t^{2}}{{a + b + (a - b) t^{2}}^{2}} d t$ $c a s e 1 a > b$ $I = 2 \int \frac{(a + b) {1 - (a - b) t^{2}}}{{(a + b)}^{2} {1 + \frac{a - b}{a + b} t^{2}}^{2}} d t$ $= \frac{2}{a + b} \int \frac{1 - \frac{a - b}{a + b} t^{2}}{{1 + \frac{a - b}{a + b} t^{2}}^{2}} d t c h a n g e m e n t$ $\sqrt{\frac{a - b}{a + b}} t = x g i v e$ $I = \frac{2}{a + b} \int \frac{1 - x^{2}}{{1 + x^{2}}^{2}} \sqrt{\frac{a + b}{a - b}} d x$ $= \frac{2}{\sqrt{a^{2} - b^{2}}} \int \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} d x b u t$ $\int \frac{1 - x^{2}}{{(1 + x^{2})}^{2}} d x = \int \frac{1 + x^{2} - 2 x^{2}}{{(1 + x^{2})}^{2}} d x$ $= \int \frac{d x}{1 + x^{2}} - \int \frac{2 x^{2}}{{(1 + x^{2})}^{2}} d x$ $= a r c t a n x - \int \frac{2 x^{2}}{{(1 + x^{2})}^{2}} d x b y p a r t s$ $u^{^{'}} = \frac{- 2 x}{{(1 + x^{2})}^{2}} a n d u = x$ $\int \frac{- 2 x}{{(1 + x^{2})}^{2}} x d x = \frac{1}{1 + x^{2}} x - \int \frac{1}{1 + x^{2}} d x$ $= \frac{x}{1 + x^{2}} - a r c t a n (x) \Rightarrow$ $I = \frac{2}{\sqrt{a^{2} - b^{2}}} {a r c t a n (x) + \frac{x}{1 + x^{2}} - a r c t a n (x)} + c$ $I = \frac{2 x}{\sqrt{a^{2} - b^{2}} (1 + x^{2})}$
Commented by math khazana by abdo last updated on 12/Jun/18
$case 2 a<b we get I = (2/(a+b)) ∫ ((1 +((b−a)/(b+a))t^2 )/({1−((b−a)/(b+a))t^2 }^2 )) dt and changement (√((b−a)/(b+a)))t =x give I = (2/(a+b)) (√((b+a)/(b−a))) ∫ ((1+x^2 )/((1−x^2 )^2 ))dx = (2/(√(b^2 −a^2 ))) ∫ ((1+x^2 )/((1−x^2 )^2 )) dx but ∫ ((1+x^2 )/((1−x^2 )^2 )) dx = ∫ ((1−x^2 +2x^2 )/((1−x^2 )^2 )) dx = ∫ (dx/(1−x^2 )) + ∫ ((2x)/((1−x^2 )^2 )) dx =ln∣((1+x)/(1−x))∣ + ∫ ((2x)/((1−x^2 )^2 )) dx by parts u^′ = ((2x)/((1−x^2 )^2 )) and v =x ⇒ ∫ ((2x)/((1−x^2 )^2 )) dx = (1/(1−x^2 )) x − ∫ (1/(1−x^2 )) dx = (x/(1−x^2 )) −ln∣((1+x)/(1−x))∣ ⇒ I = (2/(√(b^2 −a^2 ))){ ln∣((1+x)/(1−x))∣ +(x/(1−x^2 )) −ln∣((1+x)/(1−x))∣ I = ((2x)/((√(b^2 −a^2 ))(1−x^2 ))) +c$
$c a s e 2 a < b w e g e t$ $I = \frac{2}{a + b} \int \frac{1 + \frac{b - a}{b + a} t^{2}}{{1 - \frac{b - a}{b + a} t^{2}}^{2}} d t a n d c h a n g e m e n t$ $\sqrt{\frac{b - a}{b + a}} t = x g i v e$ $I = \frac{2}{a + b} \sqrt{\frac{b + a}{b - a}} \int \frac{1 + x^{2}}{{(1 - x^{2})}^{2}} d x$ $= \frac{2}{\sqrt{b^{2} - a^{2}}} \int \frac{1 + x^{2}}{{(1 - x^{2})}^{2}} d x b u t$ $\int \frac{1 + x^{2}}{{(1 - x^{2})}^{2}} d x = \int \frac{1 - x^{2} + 2 x^{2}}{{(1 - x^{2})}^{2}} d x$ $= \int \frac{d x}{1 - x^{2}} + \int \frac{2 x}{{(1 - x^{2})}^{2}} d x$ $= l n ∣ \frac{1 + x}{1 - x} ∣ + \int \frac{2 x}{{(1 - x^{2})}^{2}} d x b y p a r t s$ $u^{^{'}} = \frac{2 x}{{(1 - x^{2})}^{2}} a n d v = x \Rightarrow$ $\int \frac{2 x}{{(1 - x^{2})}^{2}} d x = \frac{1}{1 - x^{2}} x - \int \frac{1}{1 - x^{2}} d x$ $= \frac{x}{1 - x^{2}} - l n ∣ \frac{1 + x}{1 - x} ∣ \Rightarrow$ $I = \frac{2}{\sqrt{b^{2} - a^{2}}} {l n ∣ \frac{1 + x}{1 - x} ∣ + \frac{x}{1 - x^{2}} - l n ∣ \frac{1 + x}{1 - x} ∣$ $I = \frac{2 x}{\sqrt{b^{2} - a^{2}} (1 - x^{2})} + c$
Commented by math khazana by abdo last updated on 12/Jun/18

$I = \frac{2 x}{\sqrt{a^{2} - b^{2}} (1 + x^{2})} + c$
Commented by math khazana by abdo last updated on 12/Jun/18

$x = \sqrt{\frac{a - b}{a + b}} t a n (\frac{x}{2}) a t c a s e 1 a n d x = \sqrt{\frac{b - a}{b + a}} t a n (\frac{x}{2})$ $a t c a s e 2 .$
	Answered by behi83417@gmail.com last updated on 12/Jun/18

	$t g \frac{x}{2} = t \Rightarrow (1 + t^{2}) d x = 2 d t, c o s x = \frac{1 - t^{2}}{1 + t^{2}}$ $\Rightarrow I = \int \frac{a \frac{1 - t^{2}}{1 + t^{2}} + b}{{(a + b \frac{1 - t^{2}}{1 + t^{2}})}^{2}} . \frac{2 d t}{1 + t^{2}} =$ $= \int \frac{a + b - (a - b) t^{2}}{{(a + b + (a - b) t^{2})}^{2}} d t = \int \frac{m - {n t}^{2}}{{(m + {n t}^{2})}^{2}} d t =$ $= \frac{m}{n^{2}} \int \frac{d t}{{(\frac{m}{n} + t^{2})}^{2}} - \frac{1}{n} \int \frac{t^{2}}{{(\frac{m}{n} + t^{2})}^{2}} d t = {A I}_{1} + {B I}_{2}$ $1) I_{1} = \int \frac{d t}{{(\frac{m}{n} + t^{2})}^{2}} = \frac{n}{m} \int \frac{\frac{m}{n} + t^{2} - t^{2}}{{(\frac{m}{n} + t^{2})}^{2}} d t =$ $= \frac{n}{m} [\int \frac{d t}{(\frac{m}{n} + t^{2})} - \int \frac{t^{2}}{{(\frac{m}{n} + t^{2})}^{2}} d t] =$ $= \frac{n}{m} [\sqrt{\frac{n}{m}} . {t g}^{- 1} (\frac{t}{\sqrt{\frac{m}{n}}}) + {n I}_{2}]$ $I_{2} = \int \frac{t^{2}}{{(\frac{m}{n} + t^{2})}^{2}} d t = - \frac{t}{2 (t^{2} + \frac{m}{n})} + \frac{{t g}^{- 1} \frac{t}{\sqrt{\frac{m}{n}}}}{2 \sqrt{\frac{m}{n}}}$ $\Rightarrow I = \frac{m}{n^{2}} . I_{1} - \frac{1}{n} . I_{2} = \frac{m}{n^{2}} . \frac{n}{m} (\sqrt{\frac{n}{m}} . {t g}^{- 1} (\sqrt{\frac{n}{m}} . t) -$ $- \frac{n t}{2 (t^{2} + \frac{m}{n})} + \frac{1}{2} n \sqrt{\frac{n}{m}} {t g}^{- 1} (\sqrt{\frac{n}{m}} t)) -$ $- \frac{1}{n} (- \frac{t}{2 (t^{2} + \frac{m}{n})} + \frac{1}{2} \sqrt{\frac{n}{m}} . {t g}^{- 1} (\sqrt{\frac{n}{m}} t)) =$ $= \frac{1}{\sqrt{m n}} {t g}^{- 1} (\sqrt{\frac{n}{m}} t) - \frac{t}{2 (t^{2} + \frac{m}{n})} + \frac{1}{2} \sqrt{\frac{n}{m}} {t g}^{- 1} (\sqrt{\frac{n}{m}} t) -$ $- \frac{t}{2 n (t^{2} + \frac{m}{n})} + \frac{1}{2 \sqrt{m n}} {t g}^{- 1} (\sqrt{\frac{n}{m}} t) =$ $= \frac{3 + n}{2 \sqrt{m n}} {t g}^{- 1} (\sqrt{\frac{n}{m}} t) - \frac{n + 1}{n} . \frac{t}{t^{2} + \frac{m}{n}} + C =$ $= \frac{3 + a - b}{2 \sqrt{a^{2} - b^{2}}} . {t g}^{- 1} (\sqrt{\frac{a - b}{a + b}} . t g \frac{x}{2}) -$ $- (a - b + 1) . \frac{t g \frac{x}{2}}{(a - b) {t g}^{2} \frac{x}{2} + (a + b)} . ◼$
	Answered by rahul 19 last updated on 13/Jun/18

	$Guys i gave this problem just to realise$ $that a short trick exists in this form .$ $Just multiplying and dividing by$ ${cosec}^{2} x w i l l s u f f i c e .$
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