find-the-value-of-f-x-0-pi-cosx-1-2sin-2x-dx-

Question Number 34862 by a.i msup by abdo last updated on 12/May/18

f i n d t h e v a l u e o f

f (x) = \int_{0}^{π} \frac{c o s x}{1 + 2 s i n (2 x)} d x

Commented by prof Abdo imad last updated on 14/May/18

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

a l o t i n c o m p l e x e a n a l y s i s b u t i f i n d o f t e n

s o m e t h i n g t h a t w e c a n t u n d e r s t a n d a n d

e x p l a i n i t s m o r e a n d m o r e d i f f i c u l t \dots

Commented by prof Abdo imad last updated on 12/May/18

\int_{∣ z ∣= 1} φ (z) d z = \frac{- 4 π}{\sqrt{3}} {I m (\sqrt{z_{1}} + I m (\sqrt{{\bar{z}}_{1}})}

z_{1} = \frac{\sqrt{3}}{2} - \frac{i}{2} = e^{- i \frac{π}{6}} \Rightarrow \sqrt{z_{1}} = e^{- i \frac{π}{12}} \Rightarrow

I m (\sqrt{z_{1}}) = - s i n (\frac{π}{12}) a n d I m (\sqrt{{\bar{z}}_{1}}) = s i n (\frac{π}{12})

\int_{∣ z ∣= 1} φ (z) d z = 0 \Rightarrow I = 0 .

Commented by MJS last updated on 13/May/18

I {can}^{'} t follow here :

\frac{2 π i}{\sqrt{3}} (z_{1}^{\frac{1}{2}} + z_{1}^{- \frac{1}{2}} - z_{2}^{\frac{1}{2}} - z_{2}^{- \frac{1}{2}}) = \dots

because

z_{1} = \frac{\sqrt{3}}{2} - \frac{1}{2} i

\sqrt{z_{1}} = \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{6} - \sqrt{2}}{4} i

\frac{1}{\sqrt{z_{1}}} = \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{6} - \sqrt{2}}{4} i

z_{2} = - \frac{\sqrt{3}}{2} - \frac{1}{2} i

\sqrt{z_{2}} = \frac{\sqrt{6} - \sqrt{2}}{4} - \frac{\sqrt{6} + \sqrt{2}}{4} i

\frac{1}{\sqrt{z_{2}}} = \frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{6} + \sqrt{2}}{4} i

so

\frac{2 π i}{\sqrt{3}} (z_{1}^{\frac{1}{2}} + z_{1}^{- \frac{1}{2}} - z_{2}^{\frac{1}{2}} - z_{2}^{- \frac{1}{2}}) = \frac{2 π i}{\sqrt{3}} \times \sqrt{2}

{where}^{'} s the bug ?

Commented by abdo mathsup 649 cc last updated on 13/May/18

s i r M j s n o n e a d t o u s e a l g e b r i c f o r m b e c a u s e

z_{1} = \frac{\sqrt{3}}{2} - \frac{i}{2} = e^{- i \frac{π}{6}} a n d \sqrt{z_{1}} = e^{- i \frac{π}{12}}

z_{1}^{- \frac{1}{2}} = {\overset{-^{\frac{1}{2}}}{z}}_{1} \dots . .

Commented by MJS last updated on 14/May/18

z_{1}^{\frac{1}{2}} = conj (z_{1}^{- \frac{1}{2}}) = {(- z_{2})}^{\frac{1}{2}} = conj ({(- z_{2})}^{- \frac{1}{2}})

z_{1} = e^{- i \frac{π}{6}} z_{2} = e^{- i \frac{5 π}{6}}

\sqrt{z_{1}} = e^{- i \frac{π}{12}} \sqrt{z_{2}} = e^{- i \frac{5 π}{12}}

\frac{1}{\sqrt{z_{1}}} = e^{i \frac{π}{12}} \frac{1}{\sqrt{z_{2}}} = e^{i \frac{5 π}{12}}

Commented by MJS last updated on 14/May/18

(z_{1}^{\frac{1}{2}} + z_{1}^{- \frac{1}{2}} - z_{2}^{\frac{1}{2}} - z_{2}^{- \frac{1}{2}}) \neq (z_{1}^{\frac{1}{2}} - {\overset{-^{\frac{1}{2}}}{z}}_{1} + z_{1}^{- \frac{1}{2}} - {\overset{-^{- \frac{1}{2}}}{z}}_{1})

because z_{1} \neq conj (z_{2}) but z_{1} = conj (- z_{2})

sorry but I really want to understand how

the residue method works \dots

Commented by abdo mathsup 649 cc last updated on 14/May/18

s i r M j s i w i l l d e l e t m y m e t h o d u s e d i n t h i s i n t e g r a l

a n d t r y a n o t h e r b e c a u s e t h e c o n s e p t o f r e s i d u s

i s m o r e d i f f i c u l t i n s o m e c a s e s \dots .

Commented by MJS last updated on 14/May/18

Sir, I {can}^{'} t say if your method is right or

wrong, I^{'} m just trying to follow and

hopefully understand what {you}^{'} re doing .

I believe {you}^{'} re very much better in these

things than I am, since I had been studying

maths about 25 years ago and I have forgotten

many things \dots I^{'} ll be truly thankful for any

explanation you give in this or another case

Answered by tanmay.chaudhury50@gmail.com last updated on 12/May/18

l e t k = t a n x d k = {s e c}^{2} x d x

I^{'} = \int \frac{c o s x}{1 + 4 s i n x c o s x}

= \int \frac{c o s x}{1 + 4 t a n x . {c o s}^{2} x} d x

= \int \frac{c o s x}{{c o s}^{2} x ({s e c}^{2} x + 4 t a n x)} d x

= \int \frac{{s e c}^{2} x d x}{s e c x (1 + {t a n}^{2} x + 4 t a n x)}

= \int \frac{d k}{(\sqrt{1 + k^{2})} (1 + k^{2} + 4 k)}

c o n t d

Answered by math1967 last updated on 12/May/18

2 I = \underset{0}{\int^{π}} \frac{c o s x}{1 + 2 s i n (2 x)} d x + \underset{0}{\int^{π}} \frac{c o s (π - x)}{1 + 2 s i n (π - 2 x)} d x

= \underset{0}{\int^{π}} \frac{c o s x}{1 + 2 s i n (2 x)} d x - \underset{0}{\int^{π}} \frac{c o s x}{1 + 2 s i n (2 x)} d x

= 0

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