Question and Answers Forum

IntegrationQuestion and Answers: Page 317

Question Number 26125 Answers: 0 Comments: 1

Question Number 26107 Answers: 0 Comments: 0

answer to 26024 let put c= ∫_0 ^∞ cos(ax^2 )dx and c = ∫_0 ^∞ sin(ax^2 )dx ew have c−is = ∫_0 ^∞ e^(−iax^2 ) dx =2^(−1) ∫_R e^(−iax^2 ) dx and i put x^(1/2) =r(x)(notation) so 2(c−is) = ∫_R e^(−(r(ia)x)^2 ) dx and by the changement t= r(ia) x we find 2(c+is) = (r(ia))^(−1) ∫_R e^(−t^2 ) dt = r(π)/r(ia) but r(ia) =r(i) r(a) = r(a) e^ −−>2(c+is) = r(π) r(a)^(−1) e^(−iπ/4) ^) −−> c = r(2π)/_(4r(a)) and s = r(2π)/_(4r(a))

$a n s w e r t o 26024 l e t p u t c = \int_{0}^{\infty} c o s ({a x}^{2}) d x a n d c = \int_{0}^{\infty} s i n ({a x}^{2}) d x$ $e w h a v e c - i s = \int_{0}^{\infty} e^{- {i a x}^{2}} d x = 2^{- 1} \int_{R} e^{- {i a x}^{2}} d x a n d i p u t$ $x^{1 / 2} = r (x) (n o t a t i o n) s o 2 (c - i s) = \int_{R} e^{- {(r (i a) x)}^{2}} d x$ $a n d b y t h e c h a n g e m e n t t = r (i a) x w e f i n d$ $2 (c + i s) = {(r (i a))}^{- 1} \int_{R} e^{- t^{2}} d t = r (π) / r (i a) b u t$ $r (i a) = r (i) r (a) = r (a) e^{} - - > 2 (c + i s) = r (π) r {(a)}^{- 1} e^{- i π / 4}^{)}$ $- - > c = r (2 π) /_{4 r (a)} a n d s = r (2 π) /_{4 r (a)}$

Question Number 26090 Answers: 1 Comments: 0

∫((asin^3 θ+bcos^3 θ)/(sin^2 θ.cos^2 θ))dθ

$\int \frac{a \sin^{3} θ + b \cos^{3} θ}{\sin^{2} θ . \cos^{2} θ} d θ$

Question Number 26059 Answers: 0 Comments: 0

let s give n from N find the value of ∫_0 ^1 (1 +x^2 )^(n/2) sin(n arctan(x))dx

$l e t s g i v e n f r o m N f i n d t h e v a l u e o f$ $\int_{0}^{1} {(1 + x^{2})}^{n / 2} s i n (n a r c t a n (x)) d x$

Question Number 26057 Answers: 0 Comments: 1

find a integral form of L( e^(−x^2 ) ) L(f) means laplace transform of f .

$f i n d a i n t e g r a l f o r m o f L (e^{- x^{2}})$ $L (f) m e a n s l a p l a c e t r a n s f o r m o f f .$

Question Number 26055 Answers: 1 Comments: 0

calculate ∫_0 ^1 (1+t^2 )^(1/2) dt

$c a l c u l a t e \int_{0}^{1} {(1 + t^{2})}^{1 / 2} d t$

Question Number 26054 Answers: 0 Comments: 0

if (1+cos(x))^(−1) = a_0 /_2 +Σ_(n=1) ^(n=∝) a_n cos(nx)) find a_0 and a_n ...you can use fourier series.

$i f {(1 + c o s (x))}^{- 1} = a_{0} /_{2} + \sum_{n = 1}^{n =\propto} a_{n} c o s (n x))$ $f i n d a_{0} a n d a_{n} . . . y o u c a n u s e f o u r i e r$ $s e r i e s .$

Question Number 26110 Answers: 0 Comments: 0

let s give n from N find the value of ∫_0 ^π ((sin(nx))/(sinx)) dx .

$l e t s g i v e n f r o m N f i n d t h e v a l u e o f \int_{0}^{π} \frac{s i n (n x)}{s i n x} d x .$

Question Number 26024 Answers: 0 Comments: 0

e give a element from]0.∝[ find the value of ∫_0 ^∞ cos( ax^2 ) and ∫_0 ^∞ sin( ax^2 )dx.

$e g i v e a e l e m e n t f r o m] 0 . \propto [f i n d t h e v a l u e o f \int_{0}^{\infty} c o s ({a x}^{2})$ $a n d \int_{0}^{\infty} s i n ({a x}^{2}) d x .$

Question Number 26023 Answers: 0 Comments: 0

answer to question25980 key of slution we develop the foction f(x) = sin(px) at fourier serie((f 2π periodic) f(x)= Σ_(n=1) ^(n=∝) a_n sin(nx) and a_n = 2/T ∫_([T]) sin(px)sin(nx)dx (T=2π) a_n =2π^(−1) ∫_0 ^π sin(px)sin(nx)dx−>a_n = (−1)^n sin(pπ).2n π^(−1) (n^2 − p^2 )^(−1) −−>sin(px)= 2 sin(pπ).π^(−1) Σ_(n=1) ^∝ n(−1)^(n−1) (n^2 −p^2 )^(−1) sin(nx) = 2π^(−1) sin(pπ)((1^2 −p^2 )^(−1) sin (x) −2(2^2 −p^2 )^(−1) sin(2x)+...)

$a n s w e r t o q u e s t i o n 25980 k e y o f s l u t i o n w e d e v e l o p t h e$ $f o c t i o n f (x) = s i n (p x) a t f o u r i e r s e r i e ((f 2 π p e r i o d i c)$ $f (x) = \sum_{n = 1}^{n =\propto} a_{n} s i n (n x) a n d a_{n} = 2 / T \int_{[T]} s i n (p x) s i n (n x) d x (T = 2 π)$ $a_{n} = 2 π^{- 1} \int_{0}^{π} s i n (p x) s i n (n x) d x - > a_{n} = {(- 1)}^{n} s i n (p π) . 2 n π^{- 1} {(n^{2} - p^{2})}^{- 1}$ $- - > s i n (p x) = 2 s i n (p π) . π^{- 1} \sum_{n = 1}^{\propto} n {(- 1)}^{n - 1} {(n^{2} - p^{2})}^{- 1} s i n (n x)$ $= 2 π^{- 1} s i n (p π) ({(1^{2} - p^{2})}^{- 1} s i n (x) - 2 {(2^{2} - p^{2})}^{- 1} s i n (2 x) + . . .)$

Question Number 25965 Answers: 1 Comments: 0

solve the integral ∫sin^4 θdθ

$s o l v e t h e i n t e g r a l$ $\int \sin^{4} θ d θ$

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_ <>....

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

Pg 312 Pg 313 Pg 314 Pg 315 Pg 316 Pg 317 Pg 318 Pg 319 Pg 320 Pg 321

Contact: info@tinkutara.com