calculate-0-2pi-dx-2sinx-cosx-

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

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})

\dots

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