dx-3sin-x-4cos-x-

Question Number 41561 by Tawa1 last updated on 09/Aug/18

\int \frac{dx}{3 \sin (x) + 4 \cos (x)}

Commented by math khazana by abdo last updated on 09/Aug/18

c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int \frac{1}{3 \frac{2 t}{1 + t^{2}} + 4 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= 2 \int \frac{d t}{6 t + 4 - 4 t^{2}} = \int \frac{d t}{- 2 t^{2} + 3 t + 2}

= - \int \frac{d t}{2 t^{2} - 3 t - 2} l e t d e c o m p o s e

F (t) = \frac{1}{2 t^{2} - 3 t - 2}

Δ = 9 + 16 = 25 \Rightarrow t_{1} = \frac{3 + 5}{4} = 2 a n d t_{2} = \frac{3 - 5}{4} = - \frac{1}{2}

F (t) = \frac{1}{2 (t - 2) (t + \frac{1}{2})} = \frac{a}{t - 2} + \frac{b}{t + \frac{1}{2}}

a = \frac{1}{2 (2 + \frac{1}{2})} = \frac{1}{5}

b = \frac{1}{2 (\frac{1}{2} - 2)} = - \frac{1}{3} \Rightarrow F (t) = \frac{1}{5 (t - 2)} - \frac{1}{3 (t + \frac{1}{2})}

I = - \int F (t) d t = \int \frac{1}{3 (t + \frac{1}{2})} d t - \frac{1}{5} \int \frac{d t}{t - 2} + c

= \frac{1}{3} l n ∣ t + \frac{1}{2} ∣ - \frac{1}{5} l n ∣ t - 2 ∣ + c

= \frac{1}{3} l n ∣ t a n (\frac{x}{2}) + \frac{1}{2} ∣ - \frac{1}{5} l n ∣ t a n (\frac{x}{2}) - 2 ∣ + c .

Commented by Tawa1 last updated on 10/Aug/18

God bless you sir

Commented by math khazana by abdo last updated on 10/Aug/18

t h a n k y o u s i r

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} \times \frac{1}{2} d x

d x = \frac{2 d t}{1 + t^{2}}

\int \frac{2 d t}{1 + t^{2}} \times \frac{1}{3 . \frac{2 t}{1 + t^{2}} + 4 . \frac{1 - t^{2}}{1 + t^{2}}}

2 \int \frac{d t}{6 t + 4 - 4 t^{2}}

= 2 \int \frac{d t}{- 4 (t^{2} - \frac{3}{2} t - 1)}

= \frac{- 1}{2} \int \frac{d t}{t^{2} - 2 . t . \frac{3}{4} + \frac{9}{16} - \frac{9}{16} - 1}

= \frac{- 1}{2} \int \frac{d t}{{(t - \frac{3}{4})}^{2} - {(\frac{5}{4})}^{2}}

n o w u s e f o r m j l a \int \frac{d x}{x^{2} - a^{2}}

= \frac{- 1}{2} \int \frac{d t}{(t - \frac{3}{4} + \frac{5}{4}) (t - \frac{3}{4} - \frac{5}{4})}

= \frac{- 1}{2} \int \frac{d t}{(t + \frac{1}{2}) (t - 2)}

= \frac{- 1}{2} \times \frac{2}{5} \int \frac{(t + \frac{1}{2}) - (t - 2)}{(t + \frac{1}{2}) (t - 2)} d t

= \frac{- 1}{5} {\int \frac{d t}{t - 2} - \int \frac{d t}{t + \frac{1}{2}}}

= \frac{- 1}{5} {l n (\frac{t - 2}{t + \frac{1}{2}})} + c

= \frac{- 1}{5} {l n (\frac{2 t - 4}{2 t + 1})} + c

= \frac{- 1}{5} {l n (\frac{2 t a n \frac{x}{2} - 4}{2 t a n \frac{x}{2} + 1})} + c

Commented by Tawa1 last updated on 09/Aug/18

God bless you sir

Commented by Tawa1 last updated on 09/Aug/18

Umm . i would love it if you can finish it sir . God bless you sir .

I am just learning

Commented by Tawa1 last updated on 09/Aug/18

I appreciate your effort sir . God bless you

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

a c t u a l l y i d o n o t r e m e m b e r f o r m u l a \dots b u t k n o w

h o w t h e f o r m u l a i s d e r i v e d \dots n o w i a m 48 y e a r s

b u t s t i l l i k n o w h o w t h e f o r m u l a i u s e a r e d e r i v e d

Commented by Tawa1 last updated on 09/Aug/18

Noted sir

Commented by MJS last updated on 09/Aug/18

this is called Weierstrass - Method

we have

t = \tan \frac{x}{2} \Rightarrow d t = d x \times {(\tan \frac{x}{2})}^{'} = \frac{d x}{{2 \cos}^{2} \frac{x}{2}} \Rightarrow

\Rightarrow x = 2 \arctan t; d x = 2 d t \cos^{2} \frac{x}{2} \Rightarrow

\Rightarrow \sin x = \frac{2 t}{1 + t^{2}}; \cos x = \frac{1 - t^{2}}{1 + t^{2}}; d x = \frac{2 d t}{1 + t^{2}}

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

a n o t h e r m e t h o d

l e t 3 = r c o s \propto 4 = r s i n \propto

\int \frac{d x}{r c o s \propto s i n x + r s i n \propto c o s x}

\int \frac{d x}{r {s i n (x + \propto)}}

= \frac{1}{r} \int c o s e c (x + \propto) d x

n o w u s e f o r m u l a a n d p u t v a l u e o f r s n d \propto

3^{2} + 4^{2} = r^{2} ({c o s}^{2} \propto + {s i n}^{2} \propto)

r^{2} = 25 s o r = 5

t a n \propto= \frac{4}{3} \propto= {t a n}^{- 1} (\frac{4}{3})

s o a n s i s

= \frac{1}{5} l n {t a n (\frac{x + α}{2})} + c

= \frac{1}{5} l n {t a n (\frac{x + {t a n}^{- 1} (\frac{4}{3})}{2})} + c

Commented by Tawa1 last updated on 09/Aug/18

I really appreciate sir . God bless you .

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