cos-x-5-4cos-x-2-dx-

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 = \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

\dots \dots

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} \dots \dots . 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} \dots . 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 \dots . 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

Facebook Tweet Pin