calculate I = ∫_0 ^(2π) ((cos(2x))/(cosx +sinx))dx and J =∫


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Question Number 42500 by maxmathsup by imad last updated on 26/Aug/18

$c a l c u l a t e I = \int_{0}^{2 π} \frac{c o s (2 x)}{c o s x + s i n x} d x a n d J = \int_{0}^{2 π} \frac{s i n (2 x)}{c o s x + s i n x} d x$
Commented by maxmathsup by imad last updated on 27/Aug/18
$we have I = Re( ∫_0 ^(2π) (e^(i2x) /(cosx +sinx))dx) changement e^(ix) =z give ∫_0 ^(2π) (e^(i2x) /(cosx +sonx))dx = ∫_(∣z∣=1) (z^2 /(((z+z^(−1) )/2) +((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((2z^2 )/(iz{ z+z^(−1) −i(z−z^(−1) )}))dz =∫_(∣z∣=1) ((2z^2 )/(iz^2 +i +z^2 −1))dz =∫_(∣z∣=1) ((2z^2 )/((1+i)z^2 +i−1))dz let ϕ(z) =((2z^2 )/((1+i)z^2 +i−1)) poles of ϕ? ϕ(z) =((2z^2 )/((1+i)(z^2 −((1−i)/(1+i))))) =((2z^2 )/((1+i)(z^2 −(((1−i)^2 )/2)))) =((2z^2 )/((1+i)(z^2 +i))) =((2z^2 )/((1+i)(z−(√(−i)))(z+(√(−i))))) =((2z^2 )/((1+i)(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^(−((iπ)/4)) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,e^(−((iπ)/4)) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^(−((iπ)/4)) ) =lim_(z→e^(−((iπ)/4)) ) (z−e^(−((iπ)/4)) )ϕ(z) = ((2 (−i))/((1+i)2e^(−((iπ)/4)) )) =((−i)/(1+i)) e^((iπ)/4) =((−i(1−i))/2) e^((iπ)/4) =((−1−i)/2) e^((iπ)/4) Res(ϕ,−e^(−((iπ)/4)) ) =lim_(z→−e^(−((iπ)/4)) ) (z+e^((iπ)/4) )ϕ(z) =((2(−i))/((1+i)(−2 e^(−((iπ)/4)) ))) =(i/(1+i)) e^((iπ)/4) =((i(1−i))/2) e^((iπ)/4) =((1+i)/2) e^((iπ)/4) ⇒ ∫_(∣z∣ =1) ϕ(z)dz =2iπ{((−1−i)/2) e^((iπ)/4) +((1+i)/2) e^((iπ)/4) } =0 ⇒ I =0 and J =0 .$
$w e h a v e I = R e (\int_{0}^{2 π} \frac{e^{i 2 x}}{c o s x + s i n x} d x) c h a n g e m e n t e^{i x} = z g i v e$ $\int_{0}^{2 π} \frac{e^{i 2 x}}{c o s x + s o n x} d x = \int_{∣ z ∣= 1} \frac{z^{2}}{\frac{z + z^{- 1}}{2} + \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{2 z^{2}}{i z {z + z^{- 1} - i (z - z^{- 1})}} d z = \int_{∣ z ∣= 1} \frac{2 z^{2}}{{i z}^{2} + i + z^{2} - 1} d z$ $= \int_{∣ z ∣= 1} \frac{2 z^{2}}{(1 + i) z^{2} + i - 1} d z l e t φ (z) = \frac{2 z^{2}}{(1 + i) z^{2} + i - 1} p o l e s o f φ ?$ $φ (z) = \frac{2 z^{2}}{(1 + i) (z^{2} - \frac{1 - i}{1 + i})} = \frac{2 z^{2}}{(1 + i) (z^{2} - \frac{{(1 - i)}^{2}}{2})} = \frac{2 z^{2}}{(1 + i) (z^{2} + i)}$ $= \frac{2 z^{2}}{(1 + i) (z - \sqrt{- i}) (z + \sqrt{- i})} = \frac{2 z^{2}}{(1 + i) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})} s o t h e p o l e s o f φ a r e$ $\bar{+} e^{- \frac{i π}{4}} \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, e^{- \frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$ $R e s (φ, e^{- \frac{i π}{4}}) = {l i m}_{z \to e^{- \frac{i π}{4}}} (z - e^{- \frac{i π}{4}}) φ (z) = \frac{2 (- i)}{(1 + i) 2 e^{- \frac{i π}{4}}} = \frac{- i}{1 + i} e^{\frac{i π}{4}}$ $= \frac{- i (1 - i)}{2} e^{\frac{i π}{4}} = \frac{- 1 - i}{2} e^{\frac{i π}{4}}$ $R e s (φ, - e^{- \frac{i π}{4}}) = {l i m}_{z \to - e^{- \frac{i π}{4}}} (z + e^{\frac{i π}{4}}) φ (z) = \frac{2 (- i)}{(1 + i) (- 2 e^{- \frac{i π}{4}})}$ $= \frac{i}{1 + i} e^{\frac{i π}{4}} = \frac{i (1 - i)}{2} e^{\frac{i π}{4}} = \frac{1 + i}{2} e^{\frac{i π}{4}} \Rightarrow$ $\int_{∣ z ∣ = 1} φ (z) d z = 2 i π {\frac{- 1 - i}{2} e^{\frac{i π}{4}} + \frac{1 + i}{2} e^{\frac{i π}{4}}} = 0 \Rightarrow I = 0 a n d J = 0 .$
Commented by maxmathsup by imad last updated on 27/Aug/18

$a n o t h e r w a y b u t s o e a s y w e h a v e$ $I = \int_{0}^{π} \frac{c o s (2 x)}{c o s x + s i n x} d x + \int_{π}^{2 π} \frac{c o s (2 x)}{c o s x + s i n x} d x b u t$ $\int_{π}^{2 π} \frac{c o s (2 x)}{c o s x + s i n x} d x =_{x = π + t} \int_{0}^{π} \frac{c o s (2 t)}{- c o s t - s i n t} d t = - \int_{0}^{π} \frac{c o s (2 t)}{c o s t + s i n t} d t \Rightarrow$ $I = 0 t h e s a m e m e t h o d g i v e J = 0$
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