∫ ((cos x)/((5+4cos x)^2 )) dx =


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Question Number 87340 by john santu last updated on 04/Apr/20

$\int \frac{\cos x}{{(5 + 4 \cos x)}^{2}} dx =$
Commented by john santu last updated on 04/Apr/20

$dear mr Mjs . whether methods$ $besides weierstrass substitution ?$
Commented by MJS last updated on 04/Apr/20

$all other methods are super complicated$
Commented by mathmax by abdo last updated on 04/Apr/20
$I =∫ ((cosx)/((5+4cosx)^2 ))dx ⇒I =(1/(25))∫ ((cosx)/((1+(4/5)cosx)^2 ))dx let f(a) =∫ (dx/(1+acosx)) with 0<a<1 ⇒f^′ (a) =−∫ ((cosx)/((1+acosx)^2 ))dx ⇒ ∫ ((cosx)/((1+acosx)^2 ))dx =−f^′ (a) and I =−(1/(25))f^′ ((4/5)) let explicit f(a) f(a) =_(tan((x/2))=t) ∫ ((2dt)/((1+t^2 )(1+a((1−t^2 )/(1+t^2 ))))) =∫ ((2dt)/(1+t^2 +a−at^2 )) =∫ ((2dt)/((1−a)t^2 +1+a)) =(2/(1−a))∫ (dt/(t^2 +((1+a)/(1−a)))) =_(t=(√((1+a)/(1−a)))u) (2/(1−a))×((1−a)/(1+a)) ∫ (1/(1+u^2 ))×(√((1+a)/(1−a)))du =(2/(√(1−a^2 ))) arctan((√((1−a)/(1+a)))t) +C =(2/(√(1−a^2 ))) arctan((√((1−a)/(1+a)))tan((x/2))) +C ⇒ f^′ (a) =(2(1−a^2 )^(−(1/2)) )^′ arctan((√((1−a)/(1+a)))tan((x/2))) +(2/(√(1−a^2 ))) { ((((√((1−a)/(1+a)))tan((x/2))^′ )/(1+((1−a)/(1+a))tan^2 ((x/2))))} =−(−2a)(1−a^2 )^(−(3/2)) arctan((√((1−a)/(1+a)))tan((x/2))) +(2/(√(1−a^2 )))tan((x/2))×(1/(1+((1−a)/(1+a))tan^2 ((x/2))))×(((((1−a)/(1+a)))^′ )/(2(√((1−a)/(1+a))))) =2a(1−a^2 )^(−(3/2)) arctan((√((1−a)/(1+a)))tan((x/2))) +(2/(√(1−a^2 )))tan((x/2))×(1/(1+((1−a)/(1+a))tan^2 ((x/2))))×((−2)/((1+a)^2 ×2(√((1−a)/(1+a))))) I =−(1/(25))f^′ ((4/5))$
$I = \int \frac{c o s x}{{(5 + 4 c o s x)}^{2}} d x \Rightarrow I = \frac{1}{25} \int \frac{c o s x}{{(1 + \frac{4}{5} c o s x)}^{2}} d x$ $l e t f (a) = \int \frac{d x}{1 + a c o s x} w i t h 0 < a < 1 \Rightarrow f^{^{'}} (a) = - \int \frac{c o s x}{{(1 + a c o s x)}^{2}} d x \Rightarrow$ $\int \frac{c o s x}{{(1 + a c o s x)}^{2}} d x = - f^{^{'}} (a) a n d I = - \frac{1}{25} f^{^{'}} (\frac{4}{5}) l e t e x p l i c i t f (a)$ $f (a) =_{t a n (\frac{x}{2}) = t} \int \frac{2 d t}{(1 + t^{2}) (1 + a \frac{1 - t^{2}}{1 + t^{2}})}$ $= \int \frac{2 d t}{1 + t^{2} + a - {a t}^{2}} = \int \frac{2 d t}{(1 - a) t^{2} + 1 + a} = \frac{2}{1 - a} \int \frac{d t}{t^{2} + \frac{1 + a}{1 - a}}$ $=_{t = \sqrt{\frac{1 + a}{1 - a}} u} \frac{2}{1 - a} \times \frac{1 - a}{1 + a} \int \frac{1}{1 + u^{2}} \times \sqrt{\frac{1 + a}{1 - a}} d u$ $= \frac{2}{\sqrt{1 - a^{2}}} a r c t a n (\sqrt{\frac{1 - a}{1 + a}} t) + C$ $= \frac{2}{\sqrt{1 - a^{2}}} a r c t a n (\sqrt{\frac{1 - a}{1 + a}} t a n (\frac{x}{2})) + C \Rightarrow$ $f^{^{'}} (a) = {(2 {(1 - a^{2})}^{- \frac{1}{2}})}^{^{'}} a r c t a n (\sqrt{\frac{1 - a}{1 + a}} t a n (\frac{x}{2}))$ $+ \frac{2}{\sqrt{1 - a^{2}}} {\frac{(\sqrt{\frac{1 - a}{1 + a}} t a n {(\frac{x}{2})}^{^{'}}}{1 + \frac{1 - a}{1 + a} {t a n}^{2} (\frac{x}{2})}}$ $= - (- 2 a) {(1 - a^{2})}^{- \frac{3}{2}} a r c t a n (\sqrt{\frac{1 - a}{1 + a}} t a n (\frac{x}{2}))$ $+ \frac{2}{\sqrt{1 - a^{2}}} t a n (\frac{x}{2}) \times \frac{1}{1 + \frac{1 - a}{1 + a} {t a n}^{2} (\frac{x}{2})} \times \frac{{(\frac{1 - a}{1 + a})}^{^{'}}}{2 \sqrt{\frac{1 - a}{1 + a}}}$ $= 2 a {(1 - a^{2})}^{- \frac{3}{2}} a r c t a n (\sqrt{\frac{1 - a}{1 + a}} t a n (\frac{x}{2}))$ $+ \frac{2}{\sqrt{1 - a^{2}}} t a n (\frac{x}{2}) \times \frac{1}{1 + \frac{1 - a}{1 + a} {t a n}^{2} (\frac{x}{2})} \times \frac{- 2}{{(1 + a)}^{2} \times 2 \sqrt{\frac{1 - a}{1 + a}}}$ $I = - \frac{1}{25} f^{^{'}} (\frac{4}{5})$
	Answered by ajfour last updated on 04/Apr/20

	$l e t \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} (\sec^{2} \frac{x}{2}) d x = d t$ $\Rightarrow d x = \frac{2 d t}{1 + t^{2}}, \cos x = \frac{1 - t^{2}}{1 + t^{2}}$ $I = \int \frac{(\frac{1 - t^{2}}{1 + t^{2}}) (\frac{2 d t}{1 + t^{2}})}{{(5 + \frac{4 - 4 t^{2}}{1 + t^{2}})}^{2}}$ $I = 2 \int \frac{1 - t^{2}}{{(9 + t^{2})}^{2}} d t$ $n o w l e t t = 3 \tan θ$ $d t = 3 \sec^{2} θ d θ$ $I = 2 \int \frac{1 - 9 \tan^{2} θ}{81 \sec^{4} θ} (3 \sec^{2} θ d θ)$ $s a y \tan θ = z$ $I = \frac{2}{27} \int \cos^{2} θ d θ - \frac{1}{3} \int \sin^{2} θ d θ$ $. . . . . .$
	Commented by jagoll last updated on 04/Apr/20

	$weierstrass substitution$
	Commented by ajfour last updated on 04/Apr/20

	$i s t h i s t h e n a m e t o t h e m e t h o d ?$
	Commented by jagoll last updated on 04/Apr/20

	$yes sir$
	Commented by john santu last updated on 04/Apr/20

	$yes this method weierstrass$ $substitution$
	Commented by ajfour last updated on 04/Apr/20

	$a l r i g h t, i {c o u l d n}^{'} t g o a l o n g a n y$ $o t h e r l i n e .$
	Answered by TANMAY PANACEA. last updated on 04/Apr/20

	$t = 5 + 4 c o s x \to c o s x = \frac{t - 5}{4}$ $\frac{1}{4} \int \frac{4 c o s x + 5 - 5}{{(5 + 4 c o s x)}^{2}} d x$ $\frac{1}{4} \int \frac{d x}{5 + 4 c o s x} - \frac{5}{4} \int \frac{d x}{{(5 + 4 c o s x)}^{2}} ★$ $l e t I_{1} = \int \frac{d x}{5 + 4 c o s x} a n d I_{2} = \int \frac{d x}{{(5 + 4 c o s x)}^{2}}$ $s o I = \frac{I_{1}}{4} - \frac{5 I_{2}}{4} . . . . . . . e q n (1)$ $p = \frac{s i n x}{5 + 4 c o s x}$ $\frac{d p}{d x} = \frac{(5 + 4 c o s x) c o s x - s i n x (- 4 s i n x)}{{(5 + 4 c o s x)}^{2}}$ $\frac{d p}{d x} = \frac{5 c o s x + 4}{{(5 + 4 c o s x)}^{2}} = \frac{\frac{5 (t - 5)}{4} + 4}{t^{2}} = \frac{5 t - 9}{4 t^{2}} = \frac{5}{4 t} - \frac{9}{4 t^{2}}$ $\frac{s i n x}{5 + 4 c o s x} = \frac{5}{4} \int \frac{d x}{5 + c o s x} - \frac{9}{4} \int \frac{d x}{{(5 + 4 c o s x)}^{2}}$ $n o w l o o k$ $\frac{s i n x}{5 + 4 c o s x} = \frac{5 I_{1}}{4} - \frac{9 I_{2}}{4} . . . . e q n (2)$ $[5 I_{1} - 4 (\frac{s i n x}{5 + c o s x})] \times \frac{1}{9} = I_{2}$ $◼ l o o k i f w e f i n d t h e v a l u e o f I_{1} = \int \frac{d x}{5 + 4 c o s x}$ $t h e n f r o m e q n (2) w e g e t v a l u e o f I_{2}$ $f i n a l y w e g e t I = \frac{I_{1}}{4} - \frac{5 I_{2}}{4} ◼$ $\int \frac{d x}{5 + 4 c o s x} = \int \frac{{s e c}^{2} \frac{x}{2}}{5 + 5 {t a n}^{2} \frac{x}{2} + 4 - 4 {t a n}^{2} \frac{x}{2}} d x = \int \frac{{s e c}^{2} \frac{x}{2}}{{t a n}^{2} \frac{x}{2} + 3^{2}}$ $= 2 \int \frac{d (t a n \frac{x}{2})}{3^{2} + {t a n}^{2} \frac{x}{2}} d x = 2 \times \frac{1}{3} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{3}) + c_{1} =$ $I_{1} = \frac{2}{3} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{3})$ $I_{2} = \frac{1}{9} [5 I_{1} - \frac{4 s i n x}{5 + 4 c o s x}]$ $Missing \left or extra \right$ $f i n a l l y$ $I = \frac{I_{1}}{4} - \frac{5 I_{2}}{4} = \frac{2}{12} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{3}) - \frac{5}{4} \times \frac{1}{9} [\frac{10}{3} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{3}) - \frac{4 s i n x}{5 + 4 c o s x}]$ $I = (\frac{2}{12} - \frac{50}{108}) {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{3}) + \frac{5}{36} (\frac{4 s i n x}{5 + 4 c o s x}) + C$
	Commented by john santu last updated on 04/Apr/20

	$waw . . . . amazing sir$
	Commented by TANMAY PANACEA. last updated on 04/Apr/20

	$t h a n k y o u s i r$
	Answered by MJS last updated on 04/Apr/20

	$Weierstrass is the fastest$ $\int \frac{\cos x}{{(5 + 4 \cos x)}^{2}} d x =$ $[t = \tan \frac{x}{2} \to d x = \frac{2}{t^{2} + 1} d t]$ $= - 2 \int \frac{t^{2} - 1}{{(t^{2} + 9)}^{2}} d t =$ $= 20 \int \frac{d t}{{(t^{2} + 9)}^{2}} - 2 \int \frac{d t}{t^{2} + 9} =$ $= \frac{10 t}{9 (t^{2} + 9)} - \frac{8}{27} \arctan \frac{t}{3} =$ $= \frac{5 \sin x}{9 (5 + 4 \cos x)} - \frac{8}{27} \arctan \frac{\tan \frac{x}{2}}{3} + C$
	Commented by jagoll last updated on 04/Apr/20

	$great sir . i like this method$
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