Show-that-0-2pi-cos-3x-5-4cos-x-dx-pi-12-

Question Number 27464 by tawa tawa last updated on 07/Jan/18

Show that : \int_{0}^{2 π} \frac{\cos (3 x)}{5 - 4 \cos (x)} dx = \frac{π}{12}

Commented by abdo imad last updated on 07/Jan/18

l e t p u t I = \int_{0}^{2 π} \frac{c o s (3 x)}{5 - 4 c o s x} d x a n d u s e t h e c h . e^{i x} = z

I = \int_{/ z / = 1} \frac{\frac{z^{3} + z^{- 3}}{2}}{5 - 4 \frac{z + z^{- 1}}{2}} \frac{d z}{i z} = \frac{1}{2} \int_{/ z / = 1} \frac{z^{3} + z^{- 3}}{i z (5 - 2 z - 2 z^{- 1})} d z

2 I = \int_{/ z / = 1} \frac{- i (z^{3} + z^{- 3})}{5 z - 2 z^{2} - 2} d z = \int_{/ z / = 1} \frac{i (z^{3} + z^{- 3})}{2 z^{2} - 5 z + 2} d z l e t c o n s i d e r

t h e c o m p l e x f u n c t i o n f (z) = \frac{i (z^{3} + z^{- 3})}{2 z^{2} - 5 z + 2} p o l e s o f f ?

2 z^{2} - 5 z + 2 = 0 Δ = 25 - 16 = 9

z_{1} = \frac{5 - 3}{4} = \frac{1}{2} a n d z_{2} = \frac{5 + 3}{4} = 2 a n d f (z) = \frac{i (z^{3} + z^{- 3})}{2 (z - \frac{1}{2}) (z - 2)}

w e s e e t h a t / z_{2 / > 1} s o

\int_{/ z / = 1} f (z) d z = 2 i π R e s (f, \frac{1}{2})

R e s (f, \frac{1}{2}) = {l i m}_{z - > \frac{1}{2}}^{} (z - \frac{1}{2}) f (z) = \frac{i (\frac{1}{8} + 8)}{2 (- \frac{3}{2})} = \frac{i \frac{65}{8}}{- 3}

= - i \frac{65}{24} \Rightarrow \int_{/ z / = 1} f (z) d z = 2 i π (- i \frac{65}{24}) = \frac{65 π}{12}

I = \frac{65 π}{24} .

Commented by abdo imad last updated on 07/Jan/18

t h i n k t h e r e i s a m i s t a k e i n t h e e x e r c i s e ? ? ?

Commented by tawa tawa last updated on 07/Jan/18

God bless you sir . But the answer is \frac{π}{12}

Commented by abdo imad last updated on 07/Jan/18

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

Commented by tawa tawa last updated on 07/Jan/18

I really appreciate sir . God bless you