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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) 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 \dots

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) \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 \dots