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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)+...)

$${answer}\:{to}\:{question}\mathrm{25980}\:{key}\:{of}\:{slution}\:{we}\:{develop}\:\:{the} \\ $$$${foction}\:{f}\left({x}\right)\:=\:{sin}\left({px}\right)\:{at}\:{fourier}\:{serie}\left(\left({f}\:\mathrm{2}\pi\:{periodic}\right)\right. \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{{n}=\propto} \:{a}_{{n}} {sin}\left({nx}\right)\:{and}\:\:{a}_{{n}} =\:\mathrm{2}/{T}\:\int_{\left[{T}\right]} {sin}\left({px}\right){sin}\left({nx}\right){dx}\:\:\:\left({T}=\mathrm{2}\pi\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${a}_{{n}} =\mathrm{2}\pi^{−\mathrm{1}} \:\int_{\mathrm{0}} ^{\pi} {sin}\left({px}\right){sin}\left({nx}\right){dx}−>{a}_{{n}} =\:\left(−\mathrm{1}\right)^{{n}} \:{sin}\left({p}\pi\right).\mathrm{2}{n}\:\pi^{−\mathrm{1}} \left({n}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$$−−>{sin}\left({px}\right)=\:\mathrm{2}\:{sin}\left({p}\pi\right).\pi^{−\mathrm{1}} \:\sum_{{n}=\mathrm{1}} ^{\propto} \:{n}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:{sin}\left({nx}\right) \\ $$$$=\:\mathrm{2}\pi^{−\mathrm{1}} \:{sin}\left({p}\pi\right)\left(\left(\mathrm{1}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} \:{sin}\:\left({x}\right)\:−\mathrm{2}\left(\mathrm{2}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)^{−\mathrm{1}} {sin}\left(\mathrm{2}{x}\right)+...\right) \\ $$

Question Number 25965    Answers: 1   Comments: 0

solve the integral ∫sin^4 θdθ

$${solve}\:{the}\:{integral} \\ $$$$\int\mathrm{sin}^{\mathrm{4}} \theta{d}\theta \\ $$

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) )

$${answer}\:{to}\:\mathrm{25955}.{we}\:{introduce}\:{the}\:{parametric}\:{function} \\ $$$${F}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}_{} } \right){t}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{after}\:{verifying}\:{that}\: \\ $$$${F}\:{is}\:{derivable}\:{on}\left[\mathrm{0}.\propto\left[\:\:{we}\:{find}\:\:\:\partial{F}/\partial{t}=\:\:\int_{\mathrm{0}} ^{\infty} \left(\:\left(\mathrm{1}+\left(\mathrm{1}+{x}^{\mathrm{2}} \right){t}\right)^{−\mathrm{1}} {dx}\right.\right.\right. \\ $$$$\partial{F}/\partial{t}=\mathrm{1}/\mathrm{2}\:\int_{{R}} \left({tx}^{\mathrm{2}} +{t}+\mathrm{1}\right)^{−\mathrm{1}} {dx}\:\:\:{we}\:{put}\:\:{f}\left({z}\right)\:=\left({tz}^{\mathrm{2}} +{z}+\mathrm{1}\right)^{−\mathrm{1}} \\ $$$${let}\:{find}\:{the}\:{poles}\:{of}\:{f}..{tz}^{\mathrm{2}} +{z}+\mathrm{1}=\mathrm{0}\:<−>\:\:{z}=+−{i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$${and}\:{the}\:{poles}\:{are}\:\:{z}_{\mathrm{0}} ={i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \:\:{and}\:\:\:{z}_{\mathrm{1}} =−{i}\left(\left({t}+\mathrm{1}\right){t}^{−\mathrm{1}} \right)^{\mathrm{1}/\mathrm{2}} \\ $$$${and}\:\:{f}\left({t}\right)\:\:=\left({t}\left({t}−{z}_{\mathrm{0}} \right)\left({t}−{z}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \:\:\:{by}\:{residus}\:{theorem} \\ $$$$ \\ $$$$\int_{{R}} {f}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{R}\left({f}.{z}_{\mathrm{0}} \right)\:\:=\mathrm{2}{i}\pi\:\left({t}\left({z}_{\mathrm{0}} −{z}_{\mathrm{1}} \right)\right)^{−\mathrm{1}} \\ $$$$=\pi\:{t}^{−\mathrm{1}/\mathrm{2}} \left({t}+\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}\:} \:\:\:\:−>\partial{F}/\partial{t}\:=\pi\:\mathrm{2}^{−\mathrm{1}} \:{t}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{t}\right)^{−\mathrm{1}/\mathrm{2}} \\ $$$$−>{F}\left({t}\right)\:\:=\pi\:\mathrm{2}^{−\mathrm{1}\:} \int_{\mathrm{0}} ^{{t}} \:\:\:{x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx}\:+\alpha \\ $$$$\alpha={F}\left(\mathrm{0}\right)=\mathrm{0}\:{and}\:\:{F}\left({t}\right)\:=\pi\mathrm{2}^{−\mathrm{1}} \int_{\mathrm{0}} ^{{t}} \:{x}^{−\mathrm{1}/\mathrm{2}} \left(\mathrm{1}+{x}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$${and}\:{by}\:{the}\:{changement}\:\:{x}^{\mathrm{1}/\mathrm{2}} ={u}\:\:\:{we}\:{find} \\ $$$$ \\ $$$${F}\left({t}\right)\:=\:\pi\:{ln}\left(\:{t}^{\mathrm{1}/\mathrm{2}} +\left(\mathrm{1}+{t}\right)^{\mathrm{1}/\mathrm{2}} \right)\:{so}\:\:\int_{\mathrm{0}} ^{\infty} {ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}={F}\left(\mathrm{1}\right)=\pi{ln}\left(\mathrm{1}+\mathrm{2}^{\mathrm{1}/\mathrm{2}} \right) \\ $$

Question Number 25955    Answers: 0   Comments: 0

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

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\int_{\mathrm{0}} ^{\infty} \:\:\boldsymbol{\mathrm{ln}}\left(\mathrm{2}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)^{−\mathrm{1}} \boldsymbol{\mathrm{dx}} \\ $$

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...

$${answer}\:{to}\:\mathrm{25824}\:\:{we}\:{have}\:{a}^{−{x}^{\mathrm{2}} } \:=\:{e}^{−{x}^{\mathrm{2}_{} } {ln}\left({a}\right)} \:\:{so}\:{for}\:{a}>\mathrm{1} \\ $$$${ln}\left({a}\right)=\left(\:\left({ln}\left({a}\right)\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} >>>>\int_{{R}} {a}^{−^{} {x}^{\mathrm{2}} } \:=\:\int_{{R}} {e}^{−\left({x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}} \right)^{\mathrm{2}} \right.} {dx} \\ $$$${and}\:{with}\:{the}\:{changement}\:\:{t}={x}\:\left({ln}\left({a}\right)^{\mathrm{1}/\mathrm{2}} \:\:>>>>{x}={t}\:\left(\:{ln}\left({a}\right)\right)^{−\mathrm{1}/\mathrm{2}} \right. \\ $$$$\:{we}\:{have}\:\:\int_{{R}} {a}^{−{x}^{\mathrm{2}} } {dx}\:\:=\:\pi^{\mathrm{1}/\mathrm{2}} \left({ln}\left({a}\right)\right)^{−\mathrm{1}/\mathrm{2}} ...{if}\:\mathrm{0}<{a}<\mathrm{1}\:{ln}\left({a}\right)<\mathrm{0} \\ $$$$\:{and}\:{the}\:{integrale}\:{is}\:{divergente}... \\ $$

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.

$${find}\:{the}\:{value}\:{of}\:{integral}\:\:\:\int_{{R}} \left({z}−{a}\right)^{−\mathrm{1}} {dz}\:\:{with}\:{a}\:{from}\:{C}\:\:{aplly} \\ $$$${this}\:{result}\:{to}\:{find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\infty} \left(\mathrm{2}\:+{x}^{\mathrm{4}_{} } \right)^{−\mathrm{1}} {dx}. \\ $$

Question Number 25865    Answers: 1   Comments: 0

∫((2+2x)/((x−1)(x^2 +1)))dx

$$\int\frac{\mathrm{2}+\mathrm{2}{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$

Question Number 25837    Answers: 1   Comments: 0

∫(x^n lnx)dx

$$\int\left({x}^{{n}} {lnx}\right){dx} \\ $$

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.

$${if}\:\:\int_{{R}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:\:=\:\:\pi^{\mathrm{1}/\mathrm{2}} \:\:\:\:{find}\:{value}\:\:{of}\:\:\:\int_{{R}} \:{a}^{−{x}^{\mathrm{2}_{} } } {dx}\:\:\:{with}\:{a}>\mathrm{0}\:{and}\:{a}\:{not}\:\mathrm{1}. \\ $$

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).

$${answer}\:{to}\:{q}\mathrm{25796}\:{f}_{} {ind}\:{the}\:{value}\:{off}\left({x}\right)=\:\int_{\mathrm{0}^{} } ^{\pi_{} } {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mathrm{0}<{x}<\mathrm{1}\:\:\:\partial{f}/\partial{x}=\:\int_{\mathrm{0}} ^{\pi} \:\:{cos}\theta\left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta=\pi{x}^{−\mathrm{1}} −{x}^{−\mathrm{1}} \int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta{and}\:{by}\:{the}\:{changeent}\:{tan}\theta={u}\:{then}\:{the}\:{changement}\:{u}=\left(\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}\right)^{−\mathrm{1}^{} } \right)^{} \:.{t}.{we}\:{find}\:\:\int_{\mathrm{0}} ^{\theta} \left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta\:\:=\pi\left(\mathrm{1}−{x}^{\mathrm{2}_{} } \right)^{−\mathrm{1}/\mathrm{2}} \\ $$$${so}\:\partial{f}/\partial{x}=\pi{x}^{−\mathrm{1}} −\pi{x}^{−\mathrm{1}} \left(\mathrm{1}−{x}^{−\mathrm{1}/\mathrm{2}} \right)\:{so}\:{f}\left({x}\right)=\:\pi\:{ln}\left({x}\right)−\pi\int^{{x}} \:{t}^{−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\mathrm{1}/\mathrm{2}} {dt} \\ $$$${but}\:\int\:^{{x}} {t}^{−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}=−\mathrm{1}.\:\mathrm{2}^{−\mathrm{1}} \:{ln}\left(\left(\mathrm{1}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} .\left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \right)^{−\mathrm{1}} \right)\right. \\ $$$${and}\:{f}\left({x}\right)=\pi{ln}\left({x}\right)+\pi.\mathrm{2}^{−\mathrm{1}} {ln}\left(\:\left(\mathrm{1}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \left(\left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \right)^{−\mathrm{1}} \:+\beta\right.\right.\right. \\ $$$$\beta={f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\pi=} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta=−\pi{ln}\left(\mathrm{2}\right)\:{so} \\ $$$$\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{xcos}\theta\right){d}\theta\:=\:\pi{lnx}\:+\pi\mathrm{2}^{−\mathrm{1}} {ln}\left(\left(\mathrm{1}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}^{} } \right)\left(\left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \right)^{−\mathrm{1}} −\pi{ln}\left(\mathrm{2}\right).\right.\right. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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

$${find}\:{the}\:{value}\:{of}\:{f}\left({x}\right)=\int_{\mathrm{0}^{} } ^{\pi} {ln}\left(\:\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mathrm{0}<{x}<\mathrm{1} \\ $$

Question Number 26952    Answers: 1   Comments: 0

Question Number 25777    Answers: 1   Comments: 0

∫(1/(sin x+cos x))dx

$$\int\frac{\mathrm{1}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{dx} \\ $$

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_ <>....

$${a}−{nser}\:{to}\:{question}\:\mathrm{25765}...{we}\:{put}\:{I}=\int_{\mathrm{0}} ^{\infty} \left({cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{and}\:{J}=\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}...{we}\:{have}\:\mathrm{2}\left({I}+{iJ}\right)=\int_{{R}} {e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}...{let}\:{f}\left({z}\right)={e}^{{ix}^{\mathrm{2}{n}} } \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {by}\:{residus}\:{therem}\:\int_{{R}} {f}\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi{Res}\left({f}.{i}\right)\:{but}\:{Res}\left({f}.{i}\right)=\:{lim}_{{z}−{i}} \left({z}−{i}\right){f}\left({z}\right)={e}^{{i}\left({i}\right)^{\mathrm{2}{n}} } =\:{e}^{{i}\left(−\mathrm{1}\right)_{} ^{{n}} } \left(\mathrm{2}{i}\right)^{−\mathrm{1}} {then}\:\int_{{R}} {f}\left({z}\right){dz}\:=\:\pi{e}^{\left(−\mathrm{1}\right)^{{n}} } =\:\pi\left(\:{cos}\left(−\mathrm{1}\right)^{{n}} \:+{isin}\left(−\mathrm{1}\right)^{{n}} \right)\:{then}\:\int_{\mathrm{0}} ^{\infty} {cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:=\pi\mathrm{2}^{−\mathrm{1}} {cos}\left(−\mathrm{1}\right)^{{n}} {and}\:\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}=\pi.\mathrm{2}^{−\mathrm{1}} {sin}\left(−\mathrm{1}\right)^{{n}_{} } <>....\right. \\ $$

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

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}\:{and}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left(\:{x}^{\mathrm{2}{n}} \right)^{} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx} \\ $$

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..

$${find}\:{the}\:{value}\:{of}\int_{\mathrm{0}} ^{\infty} \:{artan}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}_{} } \:\:{the}\:{key}\:{of}\:{slution}\:{put}\:{F}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:{artan}\left({xt}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \:\:\:{find}\:\partial{F}/\partial{t}\:{first}\:{then}\:{F}\left({t}\right)\:{and}\:{take}\:{t}=\mathrm{2}\:{you}\:{will}\:{of}\:{find}\:{find}\:{the}\:{value}\:{of}\:{integral}.. \\ $$

Question Number 25683    Answers: 0   Comments: 0

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

$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}^{{n}} \:\right)\left(\:\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx} \\ $$

Question Number 25682    Answers: 1   Comments: 0

we give ∫_0 ^∞ t^(a−1) (1 + t)^(−1) dt =π (sin(πa))^(−1) with 0<a<1 find the value of ∫_0 ^∞ (1 +x^(16) )^(−1) dx

$${we}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:{t}^{{a}−\mathrm{1}} \left(\mathrm{1}\:+\:{t}\right)^{−\mathrm{1}} {dt}\:=\pi\:\left({sin}\left(\pi{a}\right)\right)^{−\mathrm{1}} \:{with}\:\mathrm{0}<{a}<\mathrm{1}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}\:+{x}^{\mathrm{16}} \right)^{−\mathrm{1}} {dx} \\ $$

Question Number 25677    Answers: 0   Comments: 0

let 0<x<1 find the value of F(x) = ∫ ln (1+x cost)dt fromt=0 to t=pi

$${let}\:\mathrm{0}<{x}<\mathrm{1}\:\:{find}\:{the}\:{value}\:{of}\:{F}\left({x}\right)\:=\:\int\:\mathrm{ln}\:\left(\mathrm{1}+{x}\:{cost}\right){dt}\:{fromt}=\mathrm{0}\:{to}\:{t}={pi} \\ $$

Question Number 25676    Answers: 0   Comments: 0

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

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\boldsymbol{{sin}}\left(\hat {\boldsymbol{{x}}}{n}\:\right)/\hat {{x}}\mathrm{2}\:+\:\mathrm{1}\:{dx} \\ $$

Question Number 27181    Answers: 0   Comments: 1

find the value of ∫_0 ^(∝ ) ((ln(1+t^2 ))/(1−t^2 )) dt

$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\propto\:} \frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt} \\ $$

Question Number 27180    Answers: 0   Comments: 1

find the value of ∫∫_D (x^2 /y^2 ) dxdy with D ={(x ,y)∈R^2 / 1≤x≤2 and (1/x)≤y≤ x }.

$${find}\:{the}\:{value}\:{of}\:\:\int\int_{{D}} \:\:\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\:{dxdy}\:\:{with} \\ $$$${D}\:=\left\{\left({x}\:,{y}\right)\in\mathbb{R}^{\mathrm{2}} \:/\:\:\:\mathrm{1}\leqslant{x}\leqslant\mathrm{2}\:\:{and}\:\:\frac{\mathrm{1}}{{x}}\leqslant{y}\leqslant\:{x}\:\:\right\}. \\ $$

Question Number 25660    Answers: 0   Comments: 1

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