find-0-pi-d-x-2-2x-cos-1-with-x-real-

Tinku Tara June 3, 2023 Differentiation 0 Comments

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Question Number 72908 by mathmax by abdo last updated on 04/Nov/19

find ∫_0 ^π (dθ/(x^2 −2x cosθ +1)) with x real.

f i n d \int_{0}^{π} \frac{d θ}{x^{2} - 2 x c o s θ + 1} w i t h x r e a l .

Commented by mind is power last updated on 05/Nov/19

this is True for ∣x∣<1

this is True for ∣ x ∣< 1

Commented by Tanmay chaudhury last updated on 05/Nov/19

Commented by Tanmay chaudhury last updated on 05/Nov/19

(1/(1−x^2 ))∫_0 ^π ((1−x^2 )/(x^2 −2xcosθ+1))dθ [when x<1 ] (1/(1−x^2 ))∫_0 ^π (1+2xcosθ+2x^2 cos2θ+2x^3 cos3θ+...)dθ (1/(1−x^2 ))[∣θ∣_0 ^π +others terms=0 =(π/(1−x^2 )) answer case−II when x>1 let k=(1/x) when x>1 so k<1 ∫_0 ^π (dθ/(x^2 −2xcosθ+1))dθ ∫_0 ^π (dθ/((1/k^2 )−(2/k)cosθ+1))dθ ∫_0 ^π ((k^2 dθ)/(1−2kcosθ+k^2 ))dθ [note k<1] k^2 ×(π/(1−k^2 )) =(π/((1/k^2 )−1))=so answer is =(π/(x^2 −1))

\frac{1}{1 - x^{2}} \int_{0}^{π} \frac{1 - x^{2}}{x^{2} - 2 x c o s θ + 1} d θ [w h e n x < 1]

\frac{1}{1 - x^{2}} \int_{0}^{π} (1 + 2 x c o s θ + 2 x^{2} c o s 2 θ + 2 x^{3} c o s 3 θ + \dots) d θ

\frac{1}{1 - x^{2}} [∣ θ ∣_{0}^{π} + o t h e r s t e r m s = 0

= \frac{π}{1 - x^{2}} a n s w e r

c a s e - I I w h e n x > 1

l e t k = \frac{1}{x} w h e n x > 1 s o k < 1

\int_{0}^{π} \frac{d θ}{x^{2} - 2 x c o s θ + 1} d θ

\int_{0}^{π} \frac{d θ}{\frac{1}{k^{2}} - \frac{2}{k} c o s θ + 1} d θ

\int_{0}^{π} \frac{k^{2} d θ}{1 - 2 k c o s θ + k^{2}} d θ [n o t e k < 1]

k^{2} \times \frac{π}{1 - k^{2}}

= \frac{π}{\frac{1}{k^{2}} - 1} = s o a n s w e r i s = \frac{π}{x^{2} - 1}

Commented by mathmax by abdo last updated on 05/Nov/19

let f(x)=∫_0 ^π (dθ/(x^2 −2xcosθ +1)) changement tan((θ/2))=tgive f(x)=∫_0 ^∞ ((2dt)/((1+t^2 )(x^2 −2x((1−t^2 )/(1+t^2 )) +1))) =∫_0 ^∞ ((2dt)/(x^2 (1+t^2 )−2x(1−t^2 )+1+t^2 )) =∫_0 ^∞ ((2dt)/(x^2 +x^2 t^2 −2x+2xt^2 +1+t^2 )) =∫_0 ^∞ ((2dt)/((x^2 +2x+1)t^2 +x^2 −2x+1)) =∫_0 ^∞ ((2dt)/((x+1)^2 t^2 +(x−1)^2 )) =(1/((x+1)^2 ))∫_0 ^∞ ((2dt)/(t^2 +(((x−1)/(x+1)))^2 ))( if x≠−1 and x≠1) ⇒f(x)=_(t=∣((x−1)/(x+1))∣u) (1/((x+1)^2 ))∫_0 ^∞ (2/((((x−1)/(x+1)))^2 {1+u^2 }))∣((x−1)/(x+1))∣du =(1/((x−1)^2 ))×((∣x−1∣)/(∣x+1∣))∫_0 ^∞ ((2du)/(1+u^2 )) =(1/(∣x^2 −1∣))(2×(π/2))=(π/(∣x^2 −1∣)) ⇒ f(x)=(π/(∣x^2 −1∣))

l e t f (x) = \int_{0}^{π} \frac{d θ}{x^{2} - 2 x c o s θ + 1} c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e

f (x) = \int_{0}^{\infty} \frac{2 d t}{(1 + t^{2}) (x^{2} - 2 x \frac{1 - t^{2}}{1 + t^{2}} + 1)} = \int_{0}^{\infty} \frac{2 d t}{x^{2} (1 + t^{2}) - 2 x (1 - t^{2}) + 1 + t^{2}}

= \int_{0}^{\infty} \frac{2 d t}{x^{2} + x^{2} t^{2} - 2 x + 2 {x t}^{2} + 1 + t^{2}} = \int_{0}^{\infty} \frac{2 d t}{(x^{2} + 2 x + 1) t^{2} + x^{2} - 2 x + 1}

= \int_{0}^{\infty} \frac{2 d t}{{(x + 1)}^{2} t^{2} + {(x - 1)}^{2}} = \frac{1}{{(x + 1)}^{2}} \int_{0}^{\infty} \frac{2 d t}{t^{2} + {(\frac{x - 1}{x + 1})}^{2}} (i f x \neq - 1 a n d

x \neq 1) \Rightarrow f (x) =_{t =∣ \frac{x - 1}{x + 1} ∣ u} \frac{1}{{(x + 1)}^{2}} \int_{0}^{\infty} \frac{2}{{(\frac{x - 1}{x + 1})}^{2} {1 + u^{2}}} ∣ \frac{x - 1}{x + 1} ∣ d u

= \frac{1}{{(x - 1)}^{2}} \times \frac{∣ x - 1 ∣}{∣ x + 1 ∣} \int_{0}^{\infty} \frac{2 d u}{1 + u^{2}} = \frac{1}{∣ x^{2} - 1 ∣} (2 \times \frac{π}{2}) = \frac{π}{∣ x^{2} - 1 ∣} \Rightarrow

f (x) = \frac{π}{∣ x^{2} - 1 ∣}

Commented by Tanmay chaudhury last updated on 05/Nov/19

whom you target by this remarks...reply now

w h o m y o u t a r g e t b y t h i s r e m a r k s \dots r e p l y n o w

Commented by ajfour last updated on 05/Nov/19

Great method, Tanmay Sir..

G r e a t m e t h o d, T a n m a y S i r . .

Commented by Tanmay chaudhury last updated on 05/Nov/19

Thank you sir...

T h a n k y o u s i r \dots

Commented by ajfour last updated on 05/Nov/19

sorry sir, i concluded this on a fleeting glance, forgive me..

s o r r y s i r, i c o n c l u d e d t h i s o n a

f l e e t i n g g l a n c e, f o r g i v e m e . .

Commented by Tanmay chaudhury last updated on 05/Nov/19

Commented by Tanmay chaudhury last updated on 05/Nov/19

taken from SL loney

t a k e n f r o m S L l o n e y

Commented by Tanmay chaudhury last updated on 05/Nov/19

taken from ML khanna

t a k e n f r o m M L k h a n n a

Commented by Tanmay chaudhury last updated on 05/Nov/19

Commented by Tanmay chaudhury last updated on 05/Nov/19

sir pls check i have considerd both case x<1 and x>1 pls

s i r p l s c h e c k i h a v e c o n s i d e r d b o t h c a s e

x < 1 a n d x > 1 p l s

Commented by mind is power last updated on 05/Nov/19

nice solution

nice solution

Commented by mind is power last updated on 05/Nov/19

nice sir

nice sir

Answered by mind is power last updated on 05/Nov/19

x^2 −2xcos(θ)+1 =(x−e^(iθ) )(x−e^(−iθ) ) ∫_0 ^π (dθ/((x−e^(iθ) )(x−e^(−iθ) )))=∫_0 (dθ/((x−e^(iθ) )(xe^(iθ) −1))) f(x)=∫_0 ^π (dθ/(x^2 −2xcos(θ)+1))=f(−x) juste find f in x∈R_+ z=e^(iθ) ⇒e^(iθ) =−idz ⇒∫_1 ^(−1) ((−idz)/((x−z)(xz−1)))=i∫_1 ^(−1) (dz/((z−x)(xz−1))) =i∫_1 ^(−1) (1/((x^2 −1)(z−x)))dz+((xdz)/((1−x^2 )(xz−1))) =(i/(x^2 −1)){∫_1 ^(−1) (dz/(z−x))+∫_(−1) ^1 ((xdz)/(xz−1))} =(i/(x^2 −1))(ln(−1−x)−ln(1−x)+ln(x−1)−ln(−x−1) =(i/(x^2 −1))(ln(x−1)−ln(1−x)) ln(z_1 .z_2 )=ln(z_1 )+ln(z_2 )+2inπ if x>1⇒ln(1−x)=ln(−1(x−1)=iπ+ln(x−1) if x<1 ln(x−1)=iπ+ln(1−x) ⇒ ln(x−1)−ln(1−x)=−iπ x>1 ∫_0 ^π (dθ/(x^2 −2xcos(θ)+1))= { (((π/(x^2 −1)) x>1)),((((−π)/(x^2 −1)) 0≤ x<1)) :}

x^{2} - 2 xcos (θ) + 1

= (x - e^{i θ}) (x - e^{- i θ})

\int_{0}^{π} \frac{d θ}{(x - e^{i θ}) (x - e^{- i θ})} = \int_{0} \frac{d θ}{(x - e^{i θ}) ({xe}^{i θ} - 1)}

f (x) = \int_{0}^{π} \frac{d θ}{x^{2} - 2 xcos (θ) + 1} = f (- x)

juste find f in x \in R_{+}

z = e^{i θ} \Rightarrow e^{i θ} = - idz

\Rightarrow \int_{1}^{- 1} \frac{- idz}{(x - z) (xz - 1)} = i \int_{1}^{- 1} \frac{dz}{(z - x) (xz - 1)}

= i \int_{1}^{- 1} \frac{1}{(x^{2} - 1) (z - x)} dz + \frac{xdz}{(1 - x^{2}) (xz - 1)}

= \frac{i}{x^{2} - 1} {\int_{1}^{- 1} \frac{dz}{z - x} + \int_{- 1}^{1} \frac{xdz}{xz - 1}}

= \frac{i}{x^{2} - 1} (\ln (- 1 - x) - \ln (1 - x) + \ln (x - 1) - \ln (- x - 1)

= \frac{i}{x^{2} - 1} (\ln (x - 1) - \ln (1 - x))

\ln (z_{1} . z_{2}) = \ln (z_{1}) + \ln (z_{2}) + 2 in π

if x > 1 \Rightarrow \ln (1 - x) = \ln (- 1 (x - 1) = i π + \ln (x - 1)

if x < 1 \ln (x - 1) = i π + \ln (1 - x)

\Rightarrow

\ln (x - 1) - \ln (1 - x) = - i π x > 1

\int_{0}^{π} \frac{d θ}{x^{2} - 2 xcos (θ) + 1} = {\begin{cases} \frac{π}{x^{2} - 1} x > 1 \\ \frac{- π}{x^{2} - 1} 0 ⩽ x < 1 \end{cases}

Answered by ajfour last updated on 05/Nov/19

I=∫_0 ^( π) (dθ/((x−1)^2 +4xsin^2 (θ/2))) dividing by cos^2 (θ/2) and let tan (θ/2)=t ⇒ (1/2)sec^2 (θ/2)dθ=dt ⇒ (sec^2 (θ/2))dθ=2dt I=∫^ ((2dt)/((x−1)^2 (1+t^2 )+4xt^2 )) =(1/((x+1)^2 ))∫((2dt)/(t^2 +(((x−1)/(x+1)))^2 ))=(2/(x^2 −1))tan^(−1) [((t(x+1))/(x−1))]+c =(2/(x^2 −1))((π/2)) = (π/(x^2 −1)) . I = (π/(x^2 −1)) ∙

I = \int_{0}^{π} \frac{d θ}{{(x - 1)}^{2} + 4 x \sin^{2} \frac{θ}{2}}

d i v i d i n g b y \cos^{2} \frac{θ}{2} a n d

l e t \tan \frac{θ}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{θ}{2} d θ = d t

\Rightarrow (\sec^{2} \frac{θ}{2}) d θ = 2 d t

I = \int^{} \frac{2 d t}{{(x - 1)}^{2} (1 + t^{2}) + 4 {x t}^{2}}

= \frac{1}{{(x + 1)}^{2}} \int \frac{2 d t}{t^{2} + {(\frac{x - 1}{x + 1})}^{2}} = \frac{2}{x^{2} - 1} \tan^{- 1} [\frac{t (x + 1)}{x - 1}] + c

= \frac{2}{x^{2} - 1} (\frac{π}{2}) = \frac{π}{x^{2} - 1} .

I = \frac{π}{x^{2} - 1} \cdot

Commented by mind is power last updated on 05/Nov/19

sir if ((x+1)/(x−1))<0 lim t→∞ tan^− ((((x+1}t)/(x−1)))=−(π/2)

sir if \frac{x + 1}{x - 1} < 0

\lim t \to \infty \tan^{-} ((\frac{x + 1} t}{x - 1}) = - \frac{π}{2}

Commented by ajfour last updated on 05/Nov/19

true, thanks for the light.

t r u e, t h a n k s f o r t h e l i g h t .

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