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IntegrationQuestion and Answers: Page 319

Question Number 25960 Answers: 0 Comments: 0

answer to 25955.we introduce the parametric function F(t) =∫_0 ^∞ ln(1+(1+x^2_ )t)(1+x^2 )^(−1) dx after verifying that F is derivable on[0.∝[ we find ∂F/∂t= ∫_0 ^∞ ( (1+(1+x^2 )t)^(−1) dx ∂F/∂t=1/2 ∫_R (tx^2 +t+1)^(−1) dx we put f(z) =(tz^2 +z+1)^(−1) let find the poles of f..tz^2 +z+1=0 <−> z=+−i((t+1)t^(−1) )^(1/2) and the poles are z_0 =i((t+1)t^(−1) )^(1/2) and z_1 =−i((t+1)t^(−1) )^(1/2) and f(t) =(t(t−z_0 )(t−z_1 ))^(−1) by residus theorem ∫_R f(z)dz =2iπ R(f.z_0 ) =2iπ (t(z_0 −z_1 ))^(−1) =π t^(−1/2) (t+1)^(−1/2 ) −>∂F/∂t =π 2^(−1) t^(−1/2) (1+t)^(−1/2) −>F(t) =π 2^(−1 ) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx +α α=F(0)=0 and F(t) =π2^(−1) ∫_0 ^t x^(−1/2) (1+x)^(−1/2) dx and by the changement x^(1/2) =u we find F(t) = π ln( t^(1/2) +(1+t)^(1/2) ) so ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx=F(1)=πln(1+2^(1/2) )

$a n s w e r t o 25955 . w e i n t r o d u c e t h e p a r a m e t r i c f u n c t i o n$ $F (t) = \int_{0}^{\infty} l n (1 + (1 + x^{2_{}}) t) {(1 + x^{2})}^{- 1} d x a f t e r v e r i f y i n g t h a t$ $F i s d e r i v a b l e o n [0 . \propto [w e f i n d \partial F / \partial t = \int_{0}^{\infty} ({(1 + (1 + x^{2}) t)}^{- 1} d x$ $\partial F / \partial t = 1 / 2 \int_{R} {({t x}^{2} + t + 1)}^{- 1} d x w e p u t f (z) = {({t z}^{2} + z + 1)}^{- 1}$ $l e t f i n d t h e p o l e s o f f . . {t z}^{2} + z + 1 = 0 < - > z = + - i {((t + 1) t^{- 1})}^{1 / 2}$ $a n d t h e p o l e s a r e z_{0} = i {((t + 1) t^{- 1})}^{1 / 2} a n d z_{1} = - i {((t + 1) t^{- 1})}^{1 / 2}$ $a n d f (t) = {(t (t - z_{0}) (t - z_{1}))}^{- 1} b y r e s i d u s t h e o r e m$ $\int_{R} f (z) d z = 2 i π R (f . z_{0}) = 2 i π {(t (z_{0} - z_{1}))}^{- 1}$ $= π t^{- 1 / 2} {(t + 1)}^{- 1 / 2} - > \partial F / \partial t = π 2^{- 1} t^{- 1 / 2} {(1 + t)}^{- 1 / 2}$ $- > F (t) = π 2^{- 1} \int_{0}^{t} x^{- 1 / 2} {(1 + x)}^{- 1 / 2} d x + α$ $α = F (0) = 0 a n d F (t) = π 2^{- 1} \int_{0}^{t} x^{- 1 / 2} {(1 + x)}^{- 1 / 2} d x$ $a n d b y t h e c h a n g e m e n t x^{1 / 2} = u w e f i n d$ $F (t) = π l n (t^{1 / 2} + {(1 + t)}^{1 / 2}) s o \int_{0}^{\infty} l n (2 + x^{2}) {(1 + x^{2})}^{- 1} d x = F (1) = π l n (1 + 2^{1 / 2})$

Question Number 25955 Answers: 0 Comments: 0

find the value of ∫_0 ^∞ ln(2+x^2 )(1+x^2 )^(−1) dx

$find the value of \int_{0}^{\infty} \ln (2 + x^{2}) {(1 + x^{2})}^{- 1} dx$

Question Number 25932 Answers: 0 Comments: 0

answer to 25824 we have a^(−x^2 ) = e^(−x^2_ ln(a)) so for a>1 ln(a)=( (ln(a))^(1/2) )^2 >>>>∫_R a^(−^ x^2 ) = ∫_R e^(−(x (ln(a)^(1/2) )^2 ) dx and with the changement t=x (ln(a)^(1/2) >>>>x=t ( ln(a))^(−1/2) we have ∫_R a^(−x^2 ) dx = π^(1/2) (ln(a))^(−1/2) ...if 0<a<1 ln(a)<0 and the integrale is divergente...

$a n s w e r t o 25824 w e h a v e a^{- x^{2}} = e^{- x^{2_{}} l n (a)} s o f o r a > 1$ $l n (a) = {({(l n (a))}^{1 / 2})}^{2} >>>> \int_{R} a^{-^{} x^{2}} = \int_{R} e^{- (x {(l n {(a)}^{1 / 2})}^{2}} d x$ $a n d w i t h t h e c h a n g e m e n t t = x (l n {(a)}^{1 / 2} >>>> x = t {(l n (a))}^{- 1 / 2}$ $w e h a v e \int_{R} a^{- x^{2}} d x = π^{1 / 2} {(l n (a))}^{- 1 / 2} . . . i f 0 < a < 1 l n (a) < 0$ $a n d t h e i n t e g r a l e i s d i v e r g e n t e . . .$

Question Number 25887 Answers: 1 Comments: 0

Question Number 25868 Answers: 0 Comments: 3

Question Number 25851 Answers: 0 Comments: 1

find the value of integral ∫_R (z−a)^(−1) dz with a from C aplly this result to find the value of ∫_0 ^∞ (2 +x^4_ )^(−1) dx.

$f i n d t h e v a l u e o f i n t e g r a l \int_{R} {(z - a)}^{- 1} d z w i t h a f r o m C a p l l y$ $t h i s r e s u l t t o f i n d t h e v a l u e o f \int_{0}^{\infty} {(2 + x^{4_{}})}^{- 1} d x .$

Question Number 25865 Answers: 1 Comments: 0

∫((2+2x)/((x−1)(x^2 +1)))dx

$\int \frac{2 + 2 x}{(x - 1) (x^{2} + 1)} d x$

Question Number 25837 Answers: 1 Comments: 0

∫(x^n lnx)dx

$\int (x^{n} l n x) d x$

Question Number 25824 Answers: 0 Comments: 0

if ∫_R e^(−x^2 ) dx = π^(1/2) find value of ∫_R a^(−x^2_ ) dx with a>0 and a not 1.

$i f \int_{R} e^{- x^{2}} d x = π^{1 / 2} f i n d v a l u e o f \int_{R} a^{- x^{2_{}}} d x w i t h a > 0 a n d a n o t 1 .$

Question Number 25821 Answers: 0 Comments: 0

answer to q25796 f_ ind the value off(x)= ∫_0^ ^π_ ln(1+xcosθ)dθ with 0<x<1 ∂f/∂x= ∫_0 ^π cosθ(1+xcosθ)^(−1) dθ=πx^(−1) −x^(−1) ∫_0 ^π (1+xcosθ)^(−1) dθand by the changeent tanθ=u then the changement u=((1+x)(1+x)^(−1^ ) )^ .t.we find ∫_0 ^θ (1+xcosθ)^(−1) dθ =π(1−x^2_ )^(−1/2) so ∂f/∂x=πx^(−1) −πx^(−1) (1−x^(−1/2) ) so f(x)= π ln(x)−π∫^x t^(−1) (1−t^2 )^(−1/2) dt but ∫^x t^(−1) (1−t^2 )dt=−1. 2^(−1) ln((1+(1−x^2 )^(1/2) .(1−(1−x^2 )^(1/2) )^(−1) ) and f(x)=πln(x)+π.2^(−1) ln( (1+(1−x^2 )^(1/2) ((1−(1−x^2 )^(1/2) )^(−1) +β β=f(1)=∫_0 ^(π=) ln(1+cosθ)dθ=−πln(2) so ∫_0 ^π (1+xcosθ)dθ = πlnx +π2^(−1) ln((1+(1−x^2 )^(1/2^ ) )((1−(1−x^2 )^(1/2) )^(−1) −πln(2).

$a n s w e r t o q 25796 f_{} i n d t h e v a l u e o f f (x) = \int_{0^{}}^{π_{}} l n (1 + x c o s θ) d θ w i t h 0 < x < 1 \partial f / \partial x = \int_{0}^{π} c o s θ {(1 + x c o s θ)}^{- 1} d θ = π x^{- 1} - x^{- 1} \int_{0}^{π} {(1 + x c o s θ)}^{- 1} d θ a n d b y t h e c h a n g e e n t t a n θ = u t h e n t h e c h a n g e m e n t u = {((1 + x) {(1 + x)}^{- 1^{}})}^{} . t . w e f i n d \int_{0}^{θ} {(1 + x c o s θ)}^{- 1} d θ = π {(1 - x^{2_{}})}^{- 1 / 2}$ $s o \partial f / \partial x = π x^{- 1} - π x^{- 1} (1 - x^{- 1 / 2}) s o f (x) = π l n (x) - π \int^{x} t^{- 1} {(1 - t^{2})}^{- 1 / 2} d t$ $b u t \int^{x} t^{- 1} (1 - t^{2}) d t = - 1 . 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2} . {(1 - {(1 - x^{2})}^{1 / 2})}^{- 1})$ $a n d f (x) = π l n (x) + π . 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2} ({(1 - {(1 - x^{2})}^{1 / 2})}^{- 1} + β$ $β = f (1) = \int_{0}^{π =} l n (1 + c o s θ) d θ = - π l n (2) s o$ $\int_{0}^{π} (1 + x c o s θ) d θ = π l n x + π 2^{- 1} l n ((1 + {(1 - x^{2})}^{1 / 2^{}}) ({(1 - {(1 - x^{2})}^{1 / 2})}^{- 1} - π l n (2) .$

Question Number 25811 Answers: 0 Comments: 1

Question Number 25796 Answers: 0 Comments: 0

find the value of f(x)=∫_0^ ^π ln( 1+xcosθ)dθ with 0<x<1

$f i n d t h e v a l u e o f f (x) = \int_{0^{}}^{π} l n (1 + x c o s θ) d θ w i t h 0 < x < 1$

Question Number 26952 Answers: 1 Comments: 0

Question Number 25777 Answers: 1 Comments: 0

∫(1/(sin x+cos x))dx

$\int \frac{1}{\sin x + \cos x} d x$

Question Number 25769 Answers: 0 Comments: 0

a−nser to question 25765...we put I=∫_0 ^∞ (cos(x^(2n) )(1+x^2 )^(−1) dx and J=∫_0 ^∞ sin(x^(2n) )(1+x^2 )^(−1) dx...we have 2(I+iJ)=∫_R e^(ix^(2n) ) (1+x^2 )^(−1) dx...let f(z)=e^(ix^(2n) ) (1+x^2 )^(−1) by residus therem ∫_R f(z)dz = 2iπRes(f.i) but Res(f.i)= lim_(z−i) (z−i)f(z)=e^(i(i)^(2n) ) = e^(i(−1)_ ^n ) (2i)^(−1) then ∫_R f(z)dz = πe^((−1)^n ) = π( cos(−1)^n +isin(−1)^n ) then ∫_0 ^∞ cos(x^(2n) )(1+x^2 )^(−1) dx =π2^(−1) cos(−1)^n and ∫_0 ^∞ sin(x^(2n) )(1+x^2 )^(−1) dx=π.2^(−1) sin(−1)^n_ <>....

$a - n s e r t o q u e s t i o n 25765 . . . w e p u t I = \int_{0}^{\infty} (c o s (x^{2 n}) {(1 + x^{2})}^{- 1} d x a n d J = \int_{0}^{\infty} s i n (x^{2 n}) {(1 + x^{2})}^{- 1} d x . . . w e h a v e 2 (I + i J) = \int_{R} e^{{i x}^{2 n}} {(1 + x^{2})}^{- 1} d x . . . l e t f (z) = e^{{i x}^{2 n}} {(1 + x^{2})}^{- 1} b y r e s i d u s t h e r e m \int_{R} f (z) d z = 2 i π R e s (f . i) b u t R e s (f . i) = {l i m}_{z - i} (z - i) f (z) = e^{i {(i)}^{2 n}} = e^{i {(- 1)}_{}^{n}} {(2 i)}^{- 1} t h e n \int_{R} f (z) d z = π e^{{(- 1)}^{n}} = π (c o s {(- 1)}^{n} + i s i n {(- 1)}^{n}) t h e n \int_{0}^{\infty} c o s (x^{2 n}) {(1 + x^{2})}^{- 1} d x = π 2^{- 1} c o s {(- 1)}^{n} a n d \int_{0}^{\infty} s i n (x^{2 n}) {(1 + x^{2})}^{- 1} d x = π . 2^{- 1} s i n {(- 1)}^{n_{}} <> . . . .$

Question Number 25765 Answers: 0 Comments: 0

find the value of ∫_0 ^∞ cos(x^(2n) )(1+x^2 )^(−1) dx and ∫_0 ^∞ sin( x^(2n) )^ (1+x^2 )^(−1) dx

$f i n d t h e v a l u e o f \int_{0}^{\infty} c o s (x^{2 n}) {(1 + x^{2})}^{- 1} d x a n d \int_{0}^{\infty} s i n {(x^{2 n})}^{} {(1 + x^{2})}^{- 1} d x$

Question Number 25762 Answers: 0 Comments: 0

find the value of∫_0 ^∞ artan(2x)(1+x^2 )^(−1_ ) the key of slution put F(t)=∫_0 ^∞ artan(xt)(1+x^2 )^(−1) find ∂F/∂t first then F(t) and take t=2 you will of find find the value of integral..

$f i n d t h e v a l u e o f \int_{0}^{\infty} a r t a n (2 x) {(1 + x^{2})}^{- 1_{}} t h e k e y o f s l u t i o n p u t F (t) = \int_{0}^{\infty} a r t a n (x t) {(1 + x^{2})}^{- 1} f i n d \partial F / \partial t f i r s t t h e n F (t) a n d t a k e t = 2 y o u w i l l o f f i n d f i n d t h e v a l u e o f i n t e g r a l . .$