calculate ∫_0 ^(2π) (dx/(2sinx +cosx))


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Question Number 63666 by mathmax by abdo last updated on 07/Jul/19

$c a l c u l a t e \int_{0}^{2 π} \frac{d x}{2 s i n x + c o s x}$
Commented by MJS last updated on 07/Jul/19

$= 0$ $because {we}^{'} re integrating over a full period$
Commented by mathmax by abdo last updated on 07/Jul/19
$let I =∫_0 ^(2π) (dx/(2sinx +cosx)) changement e^(ix) =z give I =∫_(∣z∣=1) (dz/(iz( 2 ((z−z^(−1) )/(2i))+((z+z^(−1) )/2)))) =∫_(∣z∣=1) (dz/(z^2 −1 +((iz^2 +i)/2))) = ∫_(∣z∣=1) ((2dz)/(2z^2 −2 +iz^2 +i)) =∫_(∣z∣=1) ((2dz)/((2+i)z^2 −2+i)) =(2/(2+i)) ∫_(∣z∣=1) (dz/(z^2 −((2−i)/(2+i)))) let W(z)=(1/(z^2 −((2−i)/(2+i)))) poles of W? ∣((2−i)/(2+i))∣=((√5)/(√5)) =1 and ((2−i)/(2+i)) =(((√5)e^(iarctan(−(1/2))) )/((√5)e^(iarctan((1/2))) )) =e^(−2i arctan((1/2))) ⇒ (√((2−i)/(2+i)))= e^(−i arctan((1/2))) ⇒W(z) =(1/((z−e^(−iarctan((1/2))) )(z+e^(−i actan((1/2))) ))) residus theorem give ∫_(∣z∣=1) W(z)dz =2iπ {Res(W,−e^(−iarctan((1/2))) ) +Res(W,e^(−iarctan((1/2))) } Res(W,−e^(−iarctan((1/2))) )=lim_(z→−e^(−iarctan((1/2))) ) (z+e^(−i arctan((1/2))) )W(z) =(1/(−2e^(−i arctan((1/2))) )) =−(1/2) e^(i arctan((1/2))) =−(1/2) e^(i((π/2)−arctan(2))) =−(i/2) e^(−iarctan(2)) ⇒ Res(W,e^(−iarctan((1/2))) ) =(1/(2e^(−iarctan((1/2))) )) =(1/2) e^(i arctan((1/2))) =(1/2) e^(i((π/2)−arctan(2))) =(i/2) e^(−iarctan(2)) ⇒ ∫_(∣z∣=1) W(z)dz =2iπ{−(i/2) e^(−iarctan(2)) +(i/2) e^(−iarctan(2)) } =0 ⇒ I =0$
$l e t I = \int_{0}^{2 π} \frac{d x}{2 s i n x + c o s x} c h a n g e m e n t e^{i x} = z g i v e$ $I = \int_{∣ z ∣= 1} \frac{d z}{i z (2 \frac{z - z^{- 1}}{2 i} + \frac{z + z^{- 1}}{2})}$ $= \int_{∣ z ∣= 1} \frac{d z}{z^{2} - 1 + \frac{{i z}^{2} + i}{2}} = \int_{∣ z ∣= 1} \frac{2 d z}{2 z^{2} - 2 + {i z}^{2} + i}$ $= \int_{∣ z ∣= 1} \frac{2 d z}{(2 + i) z^{2} - 2 + i} = \frac{2}{2 + i} \int_{∣ z ∣= 1} \frac{d z}{z^{2} - \frac{2 - i}{2 + i}}$ $l e t W (z) = \frac{1}{z^{2} - \frac{2 - i}{2 + i}} p o l e s o f W ?$ $∣ \frac{2 - i}{2 + i} ∣= \frac{\sqrt{5}}{\sqrt{5}} = 1 a n d \frac{2 - i}{2 + i} = \frac{\sqrt{5} e^{i a r c t a n (- \frac{1}{2})}}{\sqrt{5} e^{i a r c t a n (\frac{1}{2})}} = e^{- 2 i a r c t a n (\frac{1}{2})} \Rightarrow$ $\sqrt{\frac{2 - i}{2 + i}} = e^{- i a r c t a n (\frac{1}{2})} \Rightarrow W (z) = \frac{1}{(z - e^{- i a r c t a n (\frac{1}{2})}) (z + e^{- i a c t a n (\frac{1}{2})})}$ $r e s i d u s t h e o r e m g i v e$ $\int_{∣ z ∣= 1} W (z) d z = 2 i π {R e s (W, - e^{- i a r c t a n (\frac{1}{2})}) + R e s (W, e^{- i a r c t a n (\frac{1}{2})}}$ $R e s (W, - e^{- i a r c t a n (\frac{1}{2})}) = {l i m}_{z \to - e^{- i a r c t a n (\frac{1}{2})}} (z + e^{- i a r c t a n (\frac{1}{2})}) W (z)$ $= \frac{1}{- 2 e^{- i a r c t a n (\frac{1}{2})}} = - \frac{1}{2} e^{i a r c t a n (\frac{1}{2})} = - \frac{1}{2} e^{i (\frac{π}{2} - a r c t a n (2))}$ $= - \frac{i}{2} e^{- i a r c t a n (2)} \Rightarrow$ $R e s (W, e^{- i a r c t a n (\frac{1}{2})}) = \frac{1}{2 e^{- i a r c t a n (\frac{1}{2})}} = \frac{1}{2} e^{i a r c t a n (\frac{1}{2})}$ $= \frac{1}{2} e^{i (\frac{π}{2} - a r c t a n (2))} = \frac{i}{2} e^{- i a r c t a n (2)} \Rightarrow$ $\int_{∣ z ∣= 1} W (z) d z = 2 i π {- \frac{i}{2} e^{- i a r c t a n (2)} + \frac{i}{2} e^{- i a r c t a n (2)}} = 0 \Rightarrow$ $I = 0$
Commented by MJS last updated on 07/Jul/19

$2 \sin x + \cos x = \sqrt{5} \sin (x + \arctan \frac{1}{2})$ $\frac{\sqrt{5}}{5} \int \frac{d x}{\sin (x + \arctan \frac{1}{2})} = \frac{\sqrt{5}}{5} \int \frac{d t}{\sin t} =$ $= - \ln (\frac{1}{\sin t} + \frac{1}{\tan t})$ $. . .$
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