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Question Number 28706 by students last updated on 29/Jan/18

m o s t i m p o r t a n t q u e s t i o n g o r b o a r o r i i t

s o l v e t h e i n t e g r a t i o n \frac{1}{3 s i n x + 4 c o s x}

Commented by abdo imad last updated on 29/Jan/18

l e t p u t I = \int \frac{d x}{3 s i n x + 4 c o s x} a n d u s e t h e c h . t a n (\frac{x}{2}) = t

I = \int \frac{1}{\frac{6 t}{1 + t^{2}} + 4 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 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 f i n d t h e r o o t s o f

F (t) = \frac{1}{2 t^{2} - 3 t - 2} Δ = 9 - 4 (- 4) = 25 \Rightarrow t_{1} = \frac{3 + 5}{4} = 2

t_{2} = \frac{3 - 5}{4} = \frac{- 1}{2} s o F (t) = \frac{1}{2 (t - 2) (t + \frac{1}{2})} = \frac{a}{t - 2} + \frac{b}{t + \frac{1}{2}}

a = {l i m}_{t \to 2} (t - 2) F (t) = \frac{1}{2 (2 + \frac{1}{2})} = \frac{1}{5}

b = {l i m}_{t \to - \frac{1}{2}} (t + \frac{1}{2}) F (t) = \frac{1}{2 (- \frac{1}{2} - 2)} = - \frac{1}{5}

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

\int F (t) d t = \frac{1}{5} l n ∣ \frac{t - 2}{t + \frac{1}{2}} ∣ + λ a n d

I = - \frac{1}{5} l n ∣ \frac{t - 2}{t + \frac{1}{2}} ∣ + λ = - \frac{1}{5} l n ∣ \frac{t a n (\frac{x}{2}) - 2}{t a n (\frac{x}{2}) + \frac{1}{2}} ∣ + λ

Commented by students last updated on 29/Jan/18

s i r i a m c o n f u s e d p l e a s e s i r s o l v e t h e q u e s t i o n o t h e r m e t h o d

Commented by Tinkutara last updated on 29/Jan/18

This is a similar question. Just ignore the limits and change the values.

Commented by Tinkutara last updated on 29/Jan/18

Commented by Tinkutara last updated on 29/Jan/18

Commented by students last updated on 29/Jan/18

t h a n k u s i r

Commented by abdo imad last updated on 29/Jan/18

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

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

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

b o r n s \dots

Answered by ajfour last updated on 29/Jan/18

\int \frac{d x}{3 \sin x + 4 \cos x} = \int \frac{d x}{5 \sin (x + α)}

w h e r e \cos α = \frac{3}{5}, \sin α = \frac{4}{5}

= \frac{1}{5} \int cosec (x + α) d x

= \frac{1}{5} \ln ∣ cosec (x + α) - \cot (x + α) ∣ + c

= \frac{1}{5} \ln ∣ \frac{5 - 3 \cos x + 4 \sin x}{3 \sin x + 4 \cos x} ∣ + c

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