calculate-I-0-2pi-cos-2x-cosx-sinx-dx-and-J-0-2pi-sin-2x-cosx-sinx-dx-

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

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