calculate interms of n A_n = ∫_0 ^(2π) ((cos(nx))/(cosx +sinx))dx and B_n = ∫


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Question Number 39034 by maxmathsup by imad last updated on 01/Jul/18

$c a l c u l a t e i n t e r m s o f n$ $A_{n} = \int_{0}^{2 π} \frac{c o s (n x)}{c o s x + s i n x} d x a n d B_{n} = \int_{0}^{2 π} \frac{s i n (n x)}{c o s x + s i n x} d x .$
Commented by math khazana by abdo last updated on 04/Jul/18
$we have A_n =Re( ∫_0 ^(2π) (e^(inx) /(cosx +sinx))dx)and B_n =Im( ∫_0 ^(2π) (e^(inx) /(cosx +sinx))dx) changement e^(ix) =z give ∫_0 ^(2π) (e^(inx) /(cosx +sinx))dx = ∫_(∣z∣=1) (z^n /(((z+z^(−1) )/2)+((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((−2iz^n )/(z(z+z^(−1) −i(z−z^(−1) ))))dz = ∫_(∣z∣=1) ((−2i z^n )/(z^2 +1−iz^2 +i))dz = ∫_(∣z∣=1) ((−2i z^n )/((1−i)z^2 +1+i))dz =((−2i)/(1−i)) ∫_(∣z∣=1) (z^n /(z^2 +((1+i)/(1−i)))) =((−2i(1+i))/2) ∫_(∣z∣=1) (z^n /(z^2 +i))dz =(1−i) ∫_(∣z∣=1) (z^n /(z^2 +i))dz let ϕ(z)=(z^n /(z^2 +i)) ⇒ϕ(z) =(z^n /(z^2 −((√(−i)))^2 )) = (z^n /((z−(√(−i)))(z+(√(−i))))) =(z^n /((z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^(−((iπ)/4)) ) +Res(ϕ,e^(−((iπ)/4)) )} Res(ϕ, e^(−((iπ)/4)) ) = (e^(−i((nπ)/4)) /(2e^(−i(π/4)) )) = (e^(−i(((n−1)π)/4)) /2) Res(ϕ,−e^(−((iπ)/4)) )= (((−1)^n e^(−i((nπ)/4)) )/(−2e^(−((iπ)/4)) )) =−(((−1)^n )/2) e^(−i(((n−1)π)/4)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/2) e^(−i(((n−1)π)/4)) −(((−1)^n )/2) e^(−i(((n−1)π)/4)) } =iπ(1−(−1)^n ) e^(−i(((n−1)π)/4)) =iπ(1−(−1)^n ){cos((((n−1)π)/4))−i sin((((n−1)π)/4))} =π(1−(−1)^n ){icos ((((n−1)π)/4))+sin((((n−1)π)/4))}⇒ ∫_0 ^(2π) (e^(inx) /(cosx +sinx))dx =(1−i)iπ(1−(−1)^n )e^(−i(((n−1)π)/4)) =iπ(√2)(1−(−1)^n )e^(−i(π/4)) e^(−i(((n−1)π)/4)) =iπ(√2)(1−(−1)^n ) e^(−((inπ)/4)) =π(√2) (1−(−1)^n )i{cos(((nπ)/4))−i sin(((nπ)/4))} =π(√2)(1−(−1)^n ){ i cos(((nπ)/4)) +sin(((nπ)/4))} ⇒ A_n =π(√2)(1−(−1)^n )sin(((nπ)/4)) and B_n =π(√2)(1−(−1)^n )cos(((nπ)/4)) ⇒ A_(2n) =0 and B_(2n) =0 A_(2n+1) =2π(√2)sin((((2n+1)π)/4)) and B_(2n+1) =2π(√2)cos((((2n+1)π)/4)).$
$w e h a v e A_{n} = R e (\int_{0}^{2 π} \frac{e^{i n x}}{c o s x + s i n x} d x) a n d$ $B_{n} = I m (\int_{0}^{2 π} \frac{e^{i n x}}{c o s x + s i n x} d x) c h a n g e m e n t$ $e^{i x} = z g i v e$ $\int_{0}^{2 π} \frac{e^{i n x}}{c o s x + s i n x} d x = \int_{∣ z ∣= 1} \frac{z^{n}}{\frac{z + z^{- 1}}{2} + \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{- 2 {i z}^{n}}{z (z + z^{- 1} - i (z - z^{- 1}))} d z$ $= \int_{∣ z ∣= 1} \frac{- 2 i z^{n}}{z^{2} + 1 - {i z}^{2} + i} d z$ $= \int_{∣ z ∣= 1} \frac{- 2 i z^{n}}{(1 - i) z^{2} + 1 + i} d z$ $= \frac{- 2 i}{1 - i} \int_{∣ z ∣= 1} \frac{z^{n}}{z^{2} + \frac{1 + i}{1 - i}}$ $= \frac{- 2 i (1 + i)}{2} \int_{∣ z ∣= 1} \frac{z^{n}}{z^{2} + i} d z = (1 - i) \int_{∣ z ∣= 1} \frac{z^{n}}{z^{2} + i} d z$ $l e t φ (z) = \frac{z^{n}}{z^{2} + i} \Rightarrow φ (z) = \frac{z^{n}}{z^{2} - {(\sqrt{- i})}^{2}}$ $= \frac{z^{n}}{(z - \sqrt{- i}) (z + \sqrt{- i})} = \frac{z^{n}}{(z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{- \frac{i π}{4}}) + R e s (φ, e^{- \frac{i π}{4}})}$ $R e s (φ, e^{- \frac{i π}{4}}) = \frac{e^{- i \frac{n π}{4}}}{2 e^{- i \frac{π}{4}}} = \frac{e^{- i \frac{(n - 1) π}{4}}}{2}$ $R e s (φ, - e^{- \frac{i π}{4}}) = \frac{{(- 1)}^{n} e^{- i \frac{n π}{4}}}{- 2 e^{- \frac{i π}{4}}}$ $= - \frac{{(- 1)}^{n}}{2} e^{- i \frac{(n - 1) π}{4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{1}{2} e^{- i \frac{(n - 1) π}{4}} - \frac{{(- 1)}^{n}}{2} e^{- i \frac{(n - 1) π}{4}}}$ $= i π (1 - {(- 1)}^{n}) e^{- i \frac{(n - 1) π}{4}}$ $= i π (1 - {(- 1)}^{n}) {c o s (\frac{(n - 1) π}{4}) - i s i n (\frac{(n - 1) π}{4})}$ $= π (1 - {(- 1)}^{n}) {i c o s (\frac{(n - 1) π}{4}) + s i n (\frac{(n - 1) π}{4})} \Rightarrow$ $\int_{0}^{2 π} \frac{e^{i n x}}{c o s x + s i n x} d x = (1 - i) i π (1 - {(- 1)}^{n}) e^{- i \frac{(n - 1) π}{4}}$ $= i π \sqrt{2} (1 - {(- 1)}^{n}) e^{- i \frac{π}{4}} e^{- i \frac{(n - 1) π}{4}}$ $= i π \sqrt{2} (1 - {(- 1)}^{n}) e^{- \frac{i n π}{4}}$ $= π \sqrt{2} (1 - {(- 1)}^{n}) i {c o s (\frac{n π}{4}) - i s i n (\frac{n π}{4})}$ $= π \sqrt{2} (1 - {(- 1)}^{n}) {i c o s (\frac{n π}{4}) + s i n (\frac{n π}{4})} \Rightarrow$ $A_{n} = π \sqrt{2} (1 - {(- 1)}^{n}) s i n (\frac{n π}{4}) a n d$ $B_{n} = π \sqrt{2} (1 - {(- 1)}^{n}) c o s (\frac{n π}{4}) \Rightarrow$ $A_{2 n} = 0 a n d B_{2 n} = 0$ $A_{2 n + 1} = 2 π \sqrt{2} s i n (\frac{(2 n + 1) π}{4}) a n d$ $B_{2 n + 1} = 2 π \sqrt{2} c o s (\frac{(2 n + 1) π}{4}) .$
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