1) calculate ∫_0 ^(2π) (dt/(cost +x sint)) wih x from R. 2) calculate ∫_0 ^(2π) ((sint)/((cost +xsint)^2 ))dt 3) find[the value of ∫


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Question Number 63667 by mathmax by abdo last updated on 07/Jul/19

$1) c a l c u l a t e \int_{0}^{2 π} \frac{d t}{c o s t + x s i n t} w i h x f r o m R .$ $2) c a l c u l a t e \int_{0}^{2 π} \frac{s i n t}{{(c o s t + x s i n t)}^{2}} d t$ $3) f i n d [t h e v a l u e o f \int_{0}^{2 π} \frac{d t}{c o s (2 t) + 2 s i n (2 t)}$
Commented by mathmax by abdo last updated on 07/Jul/19
$1) changement z =e^(it) give ∫_0 ^(2π) (dt/(cost +xsint)) =∫_(∣z∣=1) (dz/(iz{((z+z^(−1) )/2)+x((z−z^(−1) )/(2i))})) =∫_(∣z∣=1) (dz/(((iz^2 +i)/2)+((xz^2 −x)/2))) =∫_(∣z∣=1) ((2dz)/((x+i)z^2 −x+i)) =(2/(x+i)) ∫_(∣z∣=1) (dz/(z^2 −((x−i)/(x+i)))) we have ((x−i)/(x+i)) =(((√(x^2 +1))e^(iarctan(−(1/x))) )/((√(x^2 +1))e^(iarctan((1/x))) )) =e^(−2i arctan((1/x))) ( we suppose x≠0) ⇒(√((x−i)/(x+i)))=e^(−iarctan((1/x))) let w(z) =(1/(z^2 −((x−i)/(x+i)))) ⇒W(z)=(1/((z−e^(−iarctan((1/x))) )(z+e^(−iarctan((1/x))) ))) residus theorem give ∫_(∣z∣=1) W(z)dz =2iπ{Res(W,e^(−iarctan((1/x))) )+Res(W,−e^(−iarctan((1/x))) ) Res(W,e^(−iarctan((1/x))) )=(1/(2e^(−iarctan((1/x))) )) =(1/2) e^(iarctan((1/x))) Res(W,−e^(−iarctan((1/x))) ) =(1/(−2e^(−iarctan((1/x))) )) =−(1/2)e^(iarctan((1/x))) ⇒ the residus are opposites ⇒∫_(∣z∣=1) W(z)dz =0 ⇒ ∀x ≠0 ∫_0 ^(2π) (dt/(cost +isint)) =0 if x=0 we get ∫_0 ^(2π) (dt/(cost)) =∫_0 ^π (dt/(cost)) +∫_π ^(2π) (dt/(cost)) ∫_π ^(2π) (dt/(cost)) =_(t =π +u) ∫_0 ^π (du/(−cosu)) =−∫_0 ^π (du/(cosu)) ⇒ ∫_0 ^(2π) (dt/(cost)) =0$
$1) c h a n g e m e n t z = e^{i t} g i v e$ $\int_{0}^{2 π} \frac{d t}{c o s t + x s i n t} = \int_{∣ z ∣= 1} \frac{d z}{i z {\frac{z + z^{- 1}}{2} + x \frac{z - z^{- 1}}{2 i}}}$ $= \int_{∣ z ∣= 1} \frac{d z}{\frac{{i z}^{2} + i}{2} + \frac{{x z}^{2} - x}{2}} = \int_{∣ z ∣= 1} \frac{2 d z}{(x + i) z^{2} - x + i}$ $= \frac{2}{x + i} \int_{∣ z ∣= 1} \frac{d z}{z^{2} - \frac{x - i}{x + i}} w e h a v e \frac{x - i}{x + i} = \frac{\sqrt{x^{2} + 1} e^{i a r c t a n (- \frac{1}{x})}}{\sqrt{x^{2} + 1} e^{i a r c t a n (\frac{1}{x})}}$ $= e^{- 2 i a r c t a n (\frac{1}{x})} (w e s u p p o s e x \neq 0) \Rightarrow \sqrt{\frac{x - i}{x + i}} = e^{- i a r c t a n (\frac{1}{x})}$ $l e t w (z) = \frac{1}{z^{2} - \frac{x - i}{x + i}} \Rightarrow W (z) = \frac{1}{(z - e^{- i a r c t a n (\frac{1}{x})}) (z + e^{- i a r c t a n (\frac{1}{x})})}$ $r e s i d u s t h e o r e m g i v e$ $\int_{∣ z ∣= 1} W (z) d z = 2 i π {R e s (W, e^{- i a r c t a n (\frac{1}{x})}) + R e s (W, - e^{- i a r c t a n (\frac{1}{x})})$ $R e s (W, e^{- i a r c t a n (\frac{1}{x})}) = \frac{1}{2 e^{- i a r c t a n (\frac{1}{x})}} = \frac{1}{2} e^{i a r c t a n (\frac{1}{x})}$ $R e s (W, - e^{- i a r c t a n (\frac{1}{x})}) = \frac{1}{- 2 e^{- i a r c t a n (\frac{1}{x})}} = - \frac{1}{2} e^{i a r c t a n (\frac{1}{x})} \Rightarrow$ $t h e r e s i d u s a r e o p p o s i t e s \Rightarrow \int_{∣ z ∣= 1} W (z) d z = 0 \Rightarrow$ $\forall x \neq 0 \int_{0}^{2 π} \frac{d t}{c o s t + i s i n t} = 0$ $i f x = 0 w e g e t \int_{0}^{2 π} \frac{d t}{c o s t} = \int_{0}^{π} \frac{d t}{c o s t} + \int_{π}^{2 π} \frac{d t}{c o s t}$ $\int_{π}^{2 π} \frac{d t}{c o s t} =_{t = π + u} \int_{0}^{π} \frac{d u}{- c o s u} = - \int_{0}^{π} \frac{d u}{c o s u} \Rightarrow$ $\int_{0}^{2 π} \frac{d t}{c o s t} = 0$
Commented by mathmax by abdo last updated on 07/Jul/19

$2) l e t f (x) = \int_{0}^{2 π} \frac{d t}{c o s t + x s i n t} \Rightarrow f^{^{'}} (x) = - \int_{0}^{2 π} \frac{s i n t}{{(c o s t + x s i n t)}^{2}} d t = 0$ $(b e c a u s e f (x) = 0) \Rightarrow \int_{0}^{2 π} \frac{s i n t}{{(c o s t + x s i n t)}^{2}} d t = 0$
Commented by mathmax by abdo last updated on 07/Jul/19

$3) \int_{0}^{2 π} \frac{d t}{c o s (2 t) + 2 s i n (2 t)} =_{2 t = u} \int_{0}^{4 π} \frac{d u}{2 (c o s u + 2 s i n u)}$ $= \frac{1}{2} \int_{0}^{2 π} \frac{d u}{c o s u + 2 s i n u} + \int_{2 π}^{4 π} \frac{d u}{c o s u + 2 s i n u} = 0 b e c a u s e$ $2 π i s p e r i o d .$
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