let f(x) =∫_0 ^(2π) ((sint)/(x+sint))dt 1) calculate f(x) 2) calculate g(x) =∫_0 ^(2π) ((sint)/((x+sint)^2 )) dt 3) calculste for n∈N ∫_0 ^(2π) ((sint)/((x+sint)^n ))dt 4) calculate ∫_0 ^(2π) ((sint)/(2+sint))dt and ∫


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Question Number 54372 by maxmathsup by imad last updated on 02/Feb/19

$l e t f (x) = \int_{0}^{2 π} \frac{s i n t}{x + s i n t} d t$ $1) c a l c u l a t e f (x)$ $2) c a l c u l a t e g (x) = \int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{2}} d t$ $3) c a l c u l s t e f o r n \in N \int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{n}} d t$ $4) c a l c u l a t e \int_{0}^{2 π} \frac{s i n t}{2 + s i n t} d t a n d \int_{0}^{2 π} \frac{s i n t}{{(2 + s i n t)}^{2}} d t .$
Commented by maxmathsup by imad last updated on 03/Feb/19
$1) we have f(x)=∫_0 ^(2π) ((x+sint −x)/(x+sint))dt =2π −x ∫_0 ^(2π) (dt/(x+sint)) chang.e^(it) =z give ∫_0 ^(2π) (dt/(x+sint)) =∫_(∣z∣=1) (1/(x +((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) (dz/(iz(x+((z−z^(−1) )/(2i))))) =∫_(∣z∣=1) (dz/(ixz +((z^2 −1)/2))) =∫_(∣z∣=1) ((2dz)/(2ixz+z^2 −1)) =∫_(∣z∣=1) ((2dz)/(z^2 +2ixz −1)) let ϕ(z) =(2/(z^2 +2ixz −1)) poles of ϕ ? Δ^′ =−x^2 +1 =1−x^2 case 1 1−x^2 >0 ⇒ z_1 = −ix +(√(1−x^2 ))and z_2 =−ix−(√(1−x^2 )) ∣z_1 ∣−1 =(√(1−x^2 +x^2 ))=1−1=0 ⇒∣z_1 ∣=1 ∣z_2 ∣ −1 =(√(x^2 +1−x^2 ))−1=0 ⇒ ∣z_2 ∣=1 ∫_(∣z∣=1) ϕ(z)dz =2iπ {Res(ϕ,z_1 ) +Res(ϕ,z_2 )} but ϕ(z)=(2/((z−z_1 )(z−z_2 ))) Res(ϕ,z_1 ) =(2/(z_1 −z_2 )) =(2/(2(√(1−x^2 )))) =(1/(√(1−x^2 ))) Res(ϕ,z_2 ) = (2/(z_2 −z_1 )) =(2/(−2(√(1−x^2 )))) =((−1)/(√(1−x^2 ))) ⇒∫_(∣z∣=1) ϕ(z)dz =0 ⇒f(x)=2π case2 1−x^2 <0 ⇒∣x∣>1 ⇒Δ^′ =(i(√(x^2 −1))))^2 ⇒ z_1 =−ix +i(√(x^2 −1)) and z_2 =−ix−i(√(x^2 −1)) if x>1 ∣z_1 ∣ −1 =∣x−(√(x^2 −1))∣ −1 =x−(√(x^2 −1))−1 =x−1−(√(x^2 −1)) (x−1)^2 −(x^2 −1)=x^2 −2x +1−x^2 +1 =−2(x−1)<0 ⇒∣z_1 ∣<1 ∣z_2 ∣−1 =∣x+(√(x^2 −1))∣−1 =x−1 +(√(x^2 −1))>0 ⇒∣z_2 ∣>1 (out of circle) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) =(2/(z_1 −z_2 )) =(2/(2i(√(x^2 −1)))) =(1/(i(√(x^2 −1)))) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2π (1/(√(x^2 −1))) =((2π)/(√(x^2 −1))) ⇒f(x)=2π −((2πx)/(√(x^2 −1))) =2π{(((√(x^2 −1))−x)/(√(x^2 −1)))} .rest to study the case x<−1...$
$1) w e h a v e f (x) = \int_{0}^{2 π} \frac{x + s i n t - x}{x + s i n t} d t = 2 π - x \int_{0}^{2 π} \frac{d t}{x + s i n t} c h a n g . e^{i t} = z g i v e$ $\int_{0}^{2 π} \frac{d t}{x + s i n t} = \int_{∣ z ∣= 1} \frac{1}{x + \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z} = \int_{∣ z ∣= 1} \frac{d z}{i z (x + \frac{z - z^{- 1}}{2 i})}$ $= \int_{∣ z ∣= 1} \frac{d z}{i x z + \frac{z^{2} - 1}{2}} = \int_{∣ z ∣= 1} \frac{2 d z}{2 i x z + z^{2} - 1} = \int_{∣ z ∣= 1} \frac{2 d z}{z^{2} + 2 i x z - 1}$ $l e t φ (z) = \frac{2}{z^{2} + 2 i x z - 1} p o l e s o f φ ?$ $Δ^{^{'}} = - x^{2} + 1 = 1 - x^{2}$ $c a s e 1 1 - x^{2} > 0 \Rightarrow z_{1} = - i x + \sqrt{1 - x^{2}} a n d z_{2} = - i x - \sqrt{1 - x^{2}}$ $∣ z_{1} ∣ - 1 = \sqrt{1 - x^{2} + x^{2}} = 1 - 1 = 0 \Rightarrow∣ z_{1} ∣= 1$ $∣ z_{2} ∣ - 1 = \sqrt{x^{2} + 1 - x^{2}} - 1 = 0 \Rightarrow ∣ z_{2} ∣= 1$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, z_{1}) + R e s (φ, z_{2})} b u t φ (z) = \frac{2}{(z - z_{1}) (z - z_{2})}$ $R e s (φ, z_{1}) = \frac{2}{z_{1} - z_{2}} = \frac{2}{2 \sqrt{1 - x^{2}}} = \frac{1}{\sqrt{1 - x^{2}}}$ $R e s (φ, z_{2}) = \frac{2}{z_{2} - z_{1}} = \frac{2}{- 2 \sqrt{1 - x^{2}}} = \frac{- 1}{\sqrt{1 - x^{2}}} \Rightarrow \int_{∣ z ∣= 1} φ (z) d z = 0 \Rightarrow f (x) = 2 π$ $c a s e 2 1 - x^{2} < 0 \Rightarrow∣ x ∣> 1 \Rightarrow Δ^{^{'}} = {(i \sqrt{x^{2} - 1)})}^{2} \Rightarrow$ $z_{1} = - i x + i \sqrt{x^{2} - 1} a n d z_{2} = - i x - i \sqrt{x^{2} - 1} i f x > 1$ $∣ z_{1} ∣ - 1 =∣ x - \sqrt{x^{2} - 1} ∣ - 1 = x - \sqrt{x^{2} - 1} - 1 = x - 1 - \sqrt{x^{2} - 1}$ ${(x - 1)}^{2} - (x^{2} - 1) = x^{2} - 2 x + 1 - x^{2} + 1 = - 2 (x - 1) < 0 \Rightarrow∣ z_{1} ∣< 1$ $∣ z_{2} ∣ - 1 =∣ x + \sqrt{x^{2} - 1} ∣ - 1 = x - 1 + \sqrt{x^{2} - 1} > 0 \Rightarrow∣ z_{2} ∣> 1 (o u t o f c i r c l e)$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1})$ $R e s (φ, z_{1}) = \frac{2}{z_{1} - z_{2}} = \frac{2}{2 i \sqrt{x^{2} - 1}} = \frac{1}{i \sqrt{x^{2} - 1}} \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 2 π \frac{1}{\sqrt{x^{2} - 1}} = \frac{2 π}{\sqrt{x^{2} - 1}} \Rightarrow f (x) = 2 π - \frac{2 π x}{\sqrt{x^{2} - 1}}$ $= 2 π {\frac{\sqrt{x^{2} - 1} - x}{\sqrt{x^{2} - 1}}} . r e s t t o s t u d y t h e c a s e x < - 1 . . .$
Commented by maxmathsup by imad last updated on 03/Feb/19

$2) w e h a v e f (x) = \int_{0}^{2 π} \frac{s i n t}{x + s i n t} d t \Rightarrow f^{^{'}} (x) = - \int_{0}^{2 π} \frac{s i n t}{{(x + s i n t)}^{2}} d t = - g (x) \Rightarrow$ $g (x) = - f^{^{'}} (x)$ $c a s e 1 ∣ x ∣< 1 \Rightarrow f (x) = 2 π \Rightarrow g^{^{'}} (x) = 0$ $c a s e 2 x > 1 \Rightarrow f (x) = 2 π - 2 π \frac{x}{\sqrt{x^{2} - 1}} \Rightarrow g^{^{'}} (x) = + 2 π {(\frac{x}{\sqrt{x^{2} - 1}})}^{^{'}}$ $= 2 π \frac{\sqrt{x^{2} - 1} - x \frac{x}{\sqrt{x^{2} - 1}}}{x^{2} - 1} = 2 π \frac{x^{2} - 1 - x^{2}}{(x^{2} - 1) \sqrt{x^{2} - 1}} = - \frac{2 π}{(x^{2} - 1) \sqrt{x^{2} - 1}}$
Commented by maxmathsup by imad last updated on 03/Feb/19

$4) \int_{0}^{2 π} \frac{s i n t}{2 + s i n t} d t = f (2) = 2 π - 2 π \frac{2}{\sqrt{3}} = 2 π (1 - \frac{2}{\sqrt{3}}) = \frac{2 π (\sqrt{3} - 2)}{\sqrt{3}}$ $\int_{0}^{2 π} \frac{s i n t}{{(2 + s i n t)}^{2}} d t = g (2) = \frac{- 2 π}{3 \sqrt{3}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
	$1)∫_0 ^(2π) ((sint)/(x+sint))dt ∫((x+sint−x)/(x+sint))dt ∫dt−x∫(dt/(x+((2tan(t/2))/(1+tan^2 (t/2))))) ∫dt−x∫((sec^2 (t/2))/(xtan^2 (t/2)+x+2tan(t/2)))dt ∫dt−∫((sec^2 (t/2))/(tan^2 (t/2)+2tan(t/2)×(1/x)+(1/x^2 )+1−(1/x^2 ))) ∫dt−2∫((sec^2 (t/2)×(1/2))/((tan(t/2)+(1/x))^2 +((√(1−(1/x^2 ))) )^2 )) ∫dt−2∫((d(tan(t/2)+(1/x)))/((tan(t/2)+(1/x))^2 +((√(1−(1/x^2 ))) )^2 )) so answer is ∣t−(2/(√(1−(1/x^2 ))))tan^(−1) (((tan(t/2)+(1/x))/((√(1−(1/x^2 ))) )))∣_0 ^(2π) =[2π−{(2/(√(1−(1/x^2 ))))tan^(−1) (((tanπ+(1/x))/(√(1−(1/x^2 ))))−((tan0+(1/x))/(√(1−(1/x^2 )))))] =2π pls check mistKe if any...$
	$1) \int_{0}^{2 π} \frac{s i n t}{x + s i n t} d t$ $\int \frac{x + s i n t - x}{x + s i n t} d t$ $\int d t - x \int \frac{d t}{x + \frac{2 t a n \frac{t}{2}}{1 + {t a n}^{2} \frac{t}{2}}}$ $\int d t - x \int \frac{{s e c}^{2} \frac{t}{2}}{{x t a n}^{2} \frac{t}{2} + x + 2 t a n \frac{t}{2}} d t$ $\int d t - \int \frac{{s e c}^{2} \frac{t}{2}}{{t a n}^{2} \frac{t}{2} + 2 t a n \frac{t}{2} \times \frac{1}{x} + \frac{1}{x^{2}} + 1 - \frac{1}{x^{2}}}$ $\int d t - 2 \int \frac{{s e c}^{2} \frac{t}{2} \times \frac{1}{2}}{{(t a n \frac{t}{2} + \frac{1}{x})}^{2} + {(\sqrt{1 - \frac{1}{x^{2}}})}^{2}}$ $\int d t - 2 \int \frac{d (t a n \frac{t}{2} + \frac{1}{x})}{{(t a n \frac{t}{2} + \frac{1}{x})}^{2} + {(\sqrt{1 - \frac{1}{x^{2}}})}^{2}}$ $s o a n s w e r i s$ $∣ t - \frac{2}{\sqrt{1 - \frac{1}{x^{2}}}} {t a n}^{- 1} (\frac{t a n \frac{t}{2} + \frac{1}{x}}{\sqrt{1 - \frac{1}{x^{2}}}}) ∣_{0}^{2 π}$ $= [2 π - {\frac{2}{\sqrt{1 - \frac{1}{x^{2}}}} {t a n}^{- 1} (\frac{t a n π + \frac{1}{x}}{\sqrt{1 - \frac{1}{x^{2}}}} - \frac{t a n 0 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x^{2}}}})]$ $= 2 π$ $p l s c h e c k m i s t K e i f a n y . . .$
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