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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 25823 Answers: 0 Comments: 1

z_1 =4−3i z_2 =9+2i Find (a) z_1 oz_2 (b) z_1 ×z_2 (c) angle between z_1 and z_2 .

$${z}_{\mathrm{1}} =\mathrm{4}−\mathrm{3}{i} \\ $$$${z}_{\mathrm{2}} =\mathrm{9}+\mathrm{2}{i} \\ $$$${Find}\:\:\left({a}\right)\:{z}_{\mathrm{1}} {oz}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}\right)\:{z}_{\mathrm{1}} ×{z}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({c}\right)\:{angle}\:{between}\:{z}_{\mathrm{1}} {and}\:{z}_{\mathrm{2}} . \\ $$

Question Number 25822 Answers: 0 Comments: 0

answer to the question of p(X)= (1+iX)^n −(1−iX)^n key of solutionafter resolving p(X)=0 the roots of p(X) are x_k =tan(kπ/n) with k in [[0.n−1]] so p(X)= ∝Π_(k=0) ^(k=n−1) (X−x_k^ ) let searsh ∝ by using binome formula p(X)= 2iΣ_(p=0) ^ (−1)^(p ) C_n ^(2p+1) X^(2p+1) so ∝= 2i(−1)^([n−1/2]) C_n ^ case1 n=2N p(X)=∝ Π_(k=0) ^(k=2N−1) (X−tan(kπ/2N)) and ∝=4in(−1)^(N−1) case2 n=2N+1 p(X)=∝Π_(k=0) ^(k=2N) (X−tan(kπ/2N+1) and ∝=2i(−1)^N

$${answer}\:{to}\:{the}\:{question}\:{of}\:{p}\left({X}\right)=\:\left(\mathrm{1}+{iX}\right)^{{n}} −\left(\mathrm{1}−{iX}\right)^{{n}} \\ $$$${key}\:{of}\:{solutionafter}\:{resolving}\:{p}\left({X}\right)=\mathrm{0}\:\:{the}\:{roots}\:{of}\:{p}\left({X}\right) \\ $$$${are}\:\:{x}_{{k}} ={tan}\left({k}\pi/{n}\right)\:{with}\:{k}\:\:{in}\:\left[\left[\mathrm{0}.{n}−\mathrm{1}\right]\right]\:{so} \\ $$$${p}\left({X}\right)=\:\propto\prod_{{k}=\mathrm{0}} ^{{k}={n}−\mathrm{1}} \left({X}−{x}_{{k}^{} } \right)\:{let}\:{searsh}\:\propto\:{by}\:{using}\:{binome}\:{formula} \\ $$$${p}\left({X}\right)=\:\mathrm{2}{i}\sum_{{p}=\mathrm{0}} ^{} \left(−\mathrm{1}\right)^{{p}\:} {C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} {X}^{\mathrm{2}{p}+\mathrm{1}} {so} \\ $$$$\propto=\:\mathrm{2}{i}\left(−\mathrm{1}\right)^{\left[{n}−\mathrm{1}/\mathrm{2}\right]} \:{C}_{{n}} ^{} \\ $$$${case}\mathrm{1}\:{n}=\mathrm{2}{N}\:\:\:{p}\left({X}\right)=\propto\:\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}−\mathrm{1}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}\right)\right) \\ $$$${and}\:\propto=\mathrm{4}{in}\left(−\mathrm{1}\right)^{{N}−\mathrm{1}} \\ $$$${case}\mathrm{2}\:\:{n}=\mathrm{2}{N}+\mathrm{1}\:\:\:\:{p}\left({X}\right)=\propto\prod_{{k}=\mathrm{0}} ^{{k}=\mathrm{2}{N}} \left({X}−{tan}\left({k}\pi/\mathrm{2}{N}+\mathrm{1}\right)\right. \\ $$$${and}\:\propto=\mathrm{2}{i}\left(−\mathrm{1}\right)^{{N}} \\ $$

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).

Question Number 25814 Answers: 1 Comments: 0

find δ>0 such that ∣f(x)+1∣<0.01 when 0<∣x−2∣<δ,where f(x)=((x^2 −5x+6)/(x−2)),hence use ε_δ definition to show that ((lim)/(xtends 2))f(x)=−1

$${find}\:\delta>\mathrm{0}\:{such}\:{that}\:\mid{f}\left({x}\right)+\mathrm{1}\mid<\mathrm{0}.\mathrm{01} \\ $$$${when}\:\mathrm{0}<\mid{x}−\mathrm{2}\mid<\delta,{where} \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}{{x}−\mathrm{2}},{hence}\:{use}\:\varepsilon\_\delta\:\: \\ $$$${definition}\:{to}\:{show}\:{that}\: \\ $$$$\frac{{lim}}{{xtends}\:\mathrm{2}}{f}\left({x}\right)=−\mathrm{1} \\ $$

Question Number 25811 Answers: 0 Comments: 1

Question Number 25800 Answers: 2 Comments: 0

is sum of two periodic function is also periodic give reason

$${is}\:{sum}\:{of}\:{two}\:{periodic}\:{function}\:{is} \\ $$$${also}\:{periodic}\:{give}\:{reason} \\ $$

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 26411 Answers: 1 Comments: 0

Show that x^2 + 4xy − 2y^2 + 6x − 12y − 15 = 0 represents a pair of straight lines and that these lines together with the pair of lines x^2 + 4xy − 2y^2 = 0 form a rhombus.

$${Show}\:{that}\:{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy}\:−\:\mathrm{2}{y}^{\mathrm{2}} \:+\:\mathrm{6}{x}\:−\:\mathrm{12}{y} \\ $$$$−\:\mathrm{15}\:=\:\mathrm{0}\:{represents}\:{a}\:{pair}\:{of}\:{straight} \\ $$$${lines}\:{and}\:{that}\:{these}\:{lines}\:{together} \\ $$$${with}\:{the}\:{pair}\:{of}\:{lines}\:{x}^{\mathrm{2}} \:+\:\mathrm{4}{xy}\:−\:\mathrm{2}{y}^{\mathrm{2}} \\ $$$$=\:\mathrm{0}\:{form}\:{a}\:{rhombus}. \\ $$

Question Number 26410 Answers: 1 Comments: 0

If GCD(a,b) = 1 and GCD(c, d) = 1 a ≠ b ≠ c ≠ d ≠ 1, a < b, c < d is it possible that (a/b) + (c/d) is an integer number?

$$\mathrm{If}\:\:{GCD}\left({a},{b}\right)\:=\:\mathrm{1}\:\mathrm{and}\:{GCD}\left({c},\:{d}\right)\:=\:\mathrm{1} \\ $$$${a}\:\neq\:{b}\:\neq\:{c}\:\neq\:{d}\:\neq\:\mathrm{1},\:\:{a}\:<\:{b},\:\:{c}\:<\:{d} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{that}\:\:\frac{{a}}{{b}}\:+\:\frac{{c}}{{d}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{number}? \\ $$

Question Number 25787 Answers: 0 Comments: 2

2^x =x^(2 ) ,hence x=2 . prove

$$\mathrm{2}^{{x}} ={x}^{\mathrm{2}\:} ,{hence}\:{x}=\mathrm{2}\:.\:{prove} \\ $$

Question Number 25786 Answers: 1 Comments: 0

If C_0 + C_1 + C_2 +...+ C_n = 256, then ^(2n) C_2 is equal to

$$\mathrm{If}\:{C}_{\mathrm{0}} +\:{C}_{\mathrm{1}} +\:{C}_{\mathrm{2}} +...+\:{C}_{{n}} \:=\:\mathrm{256},\:\mathrm{then} \\ $$$$\:^{\mathrm{2}{n}} {C}_{\mathrm{2}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Question Number 25783 Answers: 1 Comments: 0

If r is a unit vector then show that ∣r×(dr/dt)∣ = ∣(dr/dt)∣

$${If}\:\:\:{r}\:\:{is}\:\:{a}\:\:{unit}\:\:{vector}\:{then}\:\:{show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mid{r}×\frac{{dr}}{{dt}}\mid\:\:=\:\:\mid\frac{{dr}}{{dt}}\mid \\ $$

Question Number 26952 Answers: 1 Comments: 0

Question Number 26951 Answers: 1 Comments: 0

Question Number 25778 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 25773 Answers: 1 Comments: 0

The least value of the function φ(x) = ∫_(5π/4) ^x (3 sin t+4 cos t) dt on the interval [ ((5π)/4), ((4π)/3) ] is

$$\mathrm{The}\:\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$$\:\:\:\phi\left({x}\right)\:=\:\underset{\mathrm{5}\pi/\mathrm{4}} {\overset{{x}} {\int}}\:\left(\mathrm{3}\:\mathrm{sin}\:{t}+\mathrm{4}\:\mathrm{cos}\:{t}\right)\:{dt} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{interval}\:\left[\:\frac{\mathrm{5}\pi}{\mathrm{4}},\:\frac{\mathrm{4}\pi}{\mathrm{3}}\:\right]\:\mathrm{is} \\ $$

Question Number 25771 Answers: 1 Comments: 0

show that ((cos 50)/(sin 40))+((sin 42)/(cos 48))−((2tan 18)/(cot 72))=0

$${show}\:{that}\:\frac{\mathrm{cos}\:\mathrm{50}}{\mathrm{sin}\:\mathrm{40}}+\frac{\mathrm{sin}\:\mathrm{42}}{\mathrm{cos}\:\mathrm{48}}−\frac{\mathrm{2tan}\:\mathrm{18}}{\mathrm{cot}\:\mathrm{72}}=\mathrm{0} \\ $$

Question Number 25770 Answers: 1 Comments: 0

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 25767 Answers: 1 Comments: 0

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 25763 Answers: 1 Comments: 0

1+cos^2 2θ=2(cos^4 θ+sin^4 θ)

$$\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}\theta=\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{sin}\:^{\mathrm{4}} \theta\right) \\ $$

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 25745 Answers: 1 Comments: 0

find the volume of the solid generated by thr revolution of the curve y(x^2 +a^2 )=a^3 about itd asymptote.

$${find}\:{the}\:{volume}\:{of}\:{the}\:{solid}\: \\ $$$${generated}\:{by}\:{thr}\:{revolution}\:{of}\:{the} \\ $$$${curve}\:{y}\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)={a}^{\mathrm{3}} \:{about}\:{itd}\: \\ $$$${asymptote}. \\ $$

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