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Question Number 38517    Answers: 2   Comments: 0

simlify A= (1/((2−(√5))^4 )) + (1/((2+(√5))^4 )) B = (1/((3−(√2))^6 )) +(1/((3+(√2))^6 ))

$${simlify} \\ $$$${A}=\:\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{4}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{4}} } \\ $$$${B}\:=\:\frac{\mathrm{1}}{\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)^{\mathrm{6}} }\:+\frac{\mathrm{1}}{\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)^{\mathrm{6}} } \\ $$

Question Number 38516    Answers: 3   Comments: 1

∫_0 ^(π/2) ∣sin x − cos x∣dx

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mid\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\mid\mathrm{d}{x} \\ $$

Question Number 38515    Answers: 1   Comments: 0

Question ; x^3 + x^3 = A) x^9 B) x^6 C) x^3 D) 1 Give a reason for your answer.

$$\:{Question}\:; \\ $$$${x}^{\mathrm{3}} \:+\:{x}^{\mathrm{3}} \:=\: \\ $$$$\left.{A}\right)\:{x}^{\mathrm{9}} \\ $$$$\left.{B}\right)\:{x}^{\mathrm{6}} \\ $$$$\left.{C}\right)\:{x}^{\mathrm{3}} \\ $$$$\left.{D}\right)\:\mathrm{1} \\ $$$${Give}\:{a}\:{reason}\:{for}\:{your}\:{answer}. \\ $$

Question Number 38492    Answers: 0   Comments: 1

Question Number 38495    Answers: 4   Comments: 0

prove that tan 3a tan 2a tan a = tan 3a − tan 2a − tan a

$${prove}\:{that} \\ $$$$\boldsymbol{\mathrm{tan}}\:\mathrm{3}\boldsymbol{{a}}\:\boldsymbol{\mathrm{tan}}\:\mathrm{2}\boldsymbol{{a}}\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{{a}}\:=\:\:\boldsymbol{\mathrm{tan}}\:\mathrm{3}\boldsymbol{{a}}\:−\:\boldsymbol{\mathrm{tan}}\:\mathrm{2}\boldsymbol{{a}}\:−\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{{a}} \\ $$

Question Number 38488    Answers: 2   Comments: 0

find the value of x if 3^x = 9x

$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{if}\: \\ $$$$\mathrm{3}^{{x}} \:=\:\mathrm{9}{x} \\ $$

Question Number 38478    Answers: 0   Comments: 1

Question Number 38477    Answers: 1   Comments: 0

Question Number 38475    Answers: 1   Comments: 0

A committee of 2 girls and 3boys is to be form from 6girls and 8boys how many different committee can be formed ?

$${A}\:{committee}\:{of}\:\mathrm{2}\:{girls}\:{and}\:\mathrm{3}{boys} \\ $$$${is}\:{to}\:{be}\:{form}\:{from}\:\mathrm{6}{girls}\:{and}\:\mathrm{8}{boys} \\ $$$${how}\:{many}\:{different}\:{committee}\:{can} \\ $$$${be}\:{formed} \\ $$$$? \\ $$

Question Number 38470    Answers: 0   Comments: 4

calculate f(t)=∫_0 ^∞ ((cos(tx))/((1+tx^2 )^2 )) dx with t≥0 2) find the values of ∫_0 ^∞ ((cos(2x))/((1+2x^2 )^2 ))dx and ∫_0 ^∞ ((cosx)/((2+x^2 )^2 ))dx

$${calculate}\:{f}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{1}+{tx}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:{with}\:{t}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{values}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cosx}}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Question Number 38469    Answers: 0   Comments: 1

calculate f(a) = ∫_(−∞) ^(+∞) ((sin(ax))/(x^2 +x+1))dx 2) find the value of ∫_(−∞) ^(+∞) ((sin(3x))/(x^2 +x+1))dx

$${calculate}\:\:{f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left({ax}\right)}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left(\mathrm{3}{x}\right)}{{x}^{\mathrm{2}} \:+{x}+\mathrm{1}}{dx} \\ $$

Question Number 38468    Answers: 0   Comments: 1

calculate ∫_(−∞) ^(+∞) ((sin(2x)sh(3x))/(4+x^2 ))dx

$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{sin}\left(\mathrm{2}{x}\right){sh}\left(\mathrm{3}{x}\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx} \\ $$

Question Number 38467    Answers: 0   Comments: 1

calculate ∫_(−∞) ^(+∞) ((cos(ax)ch(bx))/(x^2 +1))dx .

$${calculate}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{cos}\left({ax}\right){ch}\left({bx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:. \\ $$

Question Number 38466    Answers: 0   Comments: 1

let a from R find F_a (t)= ∫_(−∞) ^(+∞) ((cos(tx))/(a^2 +x^2 ))dx 2) calculate F_2 (3) and F_3 (2)

$${let}\:{a}\:{from}\:{R}\:\:{find}\:{F}_{{a}} \left({t}\right)=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{F}_{\mathrm{2}} \left(\mathrm{3}\right)\:\:{and}\:{F}_{\mathrm{3}} \left(\mathrm{2}\right) \\ $$

Question Number 38465    Answers: 0   Comments: 2

find f(x)= ∫_0 ^1 ln(1+xt^3 )dt with ∣x∣<1 . 2) calculate ∫_0 ^1 ln(1+4t^3 )dt and ∫_0 ^1 ln(2+t^3 )dt.

$${find}\:\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xt}^{\mathrm{3}} \right){dt}\:{with}\:\mid{x}\mid<\mathrm{1}\:. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+\mathrm{4}{t}^{\mathrm{3}} \right){dt}\:\:\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}+{t}^{\mathrm{3}} \right){dt}. \\ $$

Question Number 38464    Answers: 0   Comments: 1

let f(x)= ∫_0 ^1 ((ln(1−x^2 t^2 ))/t^2 )dt with ∣x∣<1 find f(x) at a simple form .

$${let}\:{f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${find}\:\:{f}\left({x}\right)\:{at}\:{a}\:{simple}\:{form}\:. \\ $$

Question Number 38463    Answers: 0   Comments: 1

calculate I = ∫_0 ^1 ((ln (1−(t^2 /4)))/t^2 )dt

$${calculate}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\:\left(\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{4}}\right)}{{t}^{\mathrm{2}} }{dt}\: \\ $$

Question Number 38462    Answers: 0   Comments: 0

find ∫_1 ^(+∞) arctan(x −(1/x))dx

$${find}\:\:\int_{\mathrm{1}} ^{+\infty} {arctan}\left({x}\:−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$

Question Number 38461    Answers: 0   Comments: 0

find f(x) = ∫_0 ^∞ arctan(1+e^(−xt) )dt with x>0 2) find ∫_0 ^∞ arctan(1+e^(−2t) )dt.

$${find}\:\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:{arctan}\left(\mathrm{1}+{e}^{−{xt}} \right){dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:{arctan}\left(\mathrm{1}+{e}^{−\mathrm{2}{t}} \right){dt}. \\ $$

Question Number 38453    Answers: 0   Comments: 4

find f(x)=∫_0 ^∞ ((1−cos(xt))/t) e^(−xt) dt with x>0 1) find asimple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(πt))/t) e^(−t) dt 3)calculate ∫_0 ^∞ ((1−cos(3t))/t) e^(−2t) dt

$${find}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}}\:{e}^{−{xt}} {dt}\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{asimple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\pi{t}\right)}{{t}}\:{e}^{−{t}} {dt} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{t}\right)}{{t}}\:{e}^{−\mathrm{2}{t}} {dt} \\ $$

Question Number 39416    Answers: 1   Comments: 0

Given that f(x) is a cubic function and f(x) = x^3 − (x^2 /4) + 5x − 7 a) find one factor of f(x) b) find (d^2 y/dx^2 ) for f(x) c) hence Evaluate y = ∫_0 ^∞ f(x).

$${Given}\:{that}\:{f}\left({x}\right)\:{is}\:{a}\:{cubic}\:{function} \\ $$$${and}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\:\mathrm{5}{x}\:−\:\mathrm{7} \\ $$$$\left.{a}\right)\:{find}\:{one}\:{factor}\:{of}\:{f}\left({x}\right) \\ $$$$\left.{b}\right)\:{find}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{for}\:{f}\left({x}\right) \\ $$$$\left.{c}\right)\:{hence}\:{Evaluate}\:\:{y}\:=\:\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right). \\ $$

Question Number 38451    Answers: 1   Comments: 1

new attempt to solve qu. 37630 ∫(dx/((√x)+(√(x+1))+(√(x+2))))= [t=x+1 → dx=dt] =∫(dt/((√(t−1))+(√t)+(√(t+1))))= [((to omit the roots)),(((√a)+(√b)+(√c) must be multiplied with)),(((−(√a)−(√b)+(√c))(−(√a)+(√b)−(√c))((√a)−(√b)−(√c)))),(((1/((√a)+(√b)+(√c)))=((a^(3/2) +b^(3/2) +c^(3/2) +2(√(abc))−((a+b)(√c)+(a+c)(√b)+(b+c)(√a)))/(a^2 +b^2 +c^2 −2(ab+ac+bc))))) ] =∫((t(√(t−1))+t(√t)+t(√(t+1))+2(√(t−1))−2(√(t+1))−2(√((t−1)t(t+1))))/(3t^2 −4))dt= =∫((t(√(t−1)))/(3t^2 −4))dt+∫((t(√t))/(3t^2 −4))dt+∫((t(√(t+1)))/(3t^2 −4))dt+2∫((√(t−1))/(3t^2 −4))dt−2∫((√(t+1))/(3t^2 −4))−2∫((√((t−1)t(t+1)))/(3t^2 −4))dt I think I can solve them all except the last one so please somebody try ∫((√((t−1)t(t+1)))/(3t^2 −4))dt=? I will do the others tomorrow

$$\mathrm{new}\:\mathrm{attempt}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{qu}.\:\mathrm{37630} \\ $$$$ \\ $$$$\int\frac{{dx}}{\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\mathrm{1}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\int\frac{{dt}}{\sqrt{{t}−\mathrm{1}}+\sqrt{{t}}+\sqrt{{t}+\mathrm{1}}}= \\ $$$$ \\ $$$$\:\:\:\:\:\begin{bmatrix}{\mathrm{to}\:\mathrm{omit}\:\mathrm{the}\:\mathrm{roots}}\\{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\:\mathrm{must}\:\mathrm{be}\:\mathrm{multiplied}\:\mathrm{with}}\\{\left(−\sqrt{{a}}−\sqrt{{b}}+\sqrt{{c}}\right)\left(−\sqrt{{a}}+\sqrt{{b}}−\sqrt{{c}}\right)\left(\sqrt{{a}}−\sqrt{{b}}−\sqrt{{c}}\right)}\\{\frac{\mathrm{1}}{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}}=\frac{{a}^{\mathrm{3}/\mathrm{2}} +{b}^{\mathrm{3}/\mathrm{2}} +{c}^{\mathrm{3}/\mathrm{2}} +\mathrm{2}\sqrt{{abc}}−\left(\left({a}+{b}\right)\sqrt{{c}}+\left({a}+{c}\right)\sqrt{{b}}+\left({b}+{c}\right)\sqrt{{a}}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}\left({ab}+{ac}+{bc}\right)}}\end{bmatrix} \\ $$$$ \\ $$$$=\int\frac{{t}\sqrt{{t}−\mathrm{1}}+{t}\sqrt{{t}}+{t}\sqrt{{t}+\mathrm{1}}+\mathrm{2}\sqrt{{t}−\mathrm{1}}−\mathrm{2}\sqrt{{t}+\mathrm{1}}−\mathrm{2}\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}= \\ $$$$=\int\frac{{t}\sqrt{{t}−\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}+\int\frac{{t}\sqrt{{t}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}+\int\frac{{t}\sqrt{{t}+\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}+\mathrm{2}\int\frac{\sqrt{{t}−\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}−\mathrm{2}\int\frac{\sqrt{{t}+\mathrm{1}}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}−\mathrm{2}\int\frac{\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{them}\:\mathrm{all}\:\mathrm{except}\:\mathrm{the}\:\mathrm{last}\:\mathrm{one} \\ $$$$\mathrm{so}\:\mathrm{please}\:\mathrm{somebody}\:\mathrm{try} \\ $$$$\int\frac{\sqrt{\left({t}−\mathrm{1}\right){t}\left({t}+\mathrm{1}\right)}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{4}}{dt}=? \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{do}\:\mathrm{the}\:\mathrm{others}\:\mathrm{tomorrow} \\ $$

Question Number 38460    Answers: 0   Comments: 1

let ∣x∣>1 find the value of F(x)=∫_0 ^∞ ln(1+xt^2 )dt 2)calculate ∫_0 ^∞ ln(1+3t^2 )dt .

$${let}\:\mid{x}\mid>\mathrm{1}\:{find}\:{the}\:{value}\:{of} \\ $$$${F}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{ln}\left(\mathrm{1}+\mathrm{3}{t}^{\mathrm{2}} \right){dt}\:. \\ $$

Question Number 38459    Answers: 0   Comments: 7

x^2 (x−b^2 )+a^2 b^2 (x−a^2 )=0 Solve for x.

$${x}^{\mathrm{2}} \left({x}−{b}^{\mathrm{2}} \right)+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({x}−{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${Solve}\:{for}\:{x}. \\ $$

Question Number 38458    Answers: 0   Comments: 3

let ∣x∣<1 calculate F(x)=∫_0 ^1 ln(1+xt^2 )dt 2) find the value of ∫_0 ^1 ln(1 +(1/2)t^2 )dt 3)find the value of A(θ) =∫_0 ^1 ln(1+sinθ t^2 )dt .

$${let}\:\mid{x}\mid<\mathrm{1}\:\:{calculate}\:\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xt}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{sin}\theta\:{t}^{\mathrm{2}} \right){dt}\:. \\ $$

Question Number 38457    Answers: 0   Comments: 0

f is a C^2 function prove that 1)L(f^′ )=x L(f)−f(0^+ ) 2)L(f^(′′) )=x^2 L(f) −xf(0^+ )−f^′ (0^+ ) L means Laplace transform.

$${f}\:{is}\:{a}\:{C}^{\mathrm{2}} {function}\:\:{prove}\:{that}\: \\ $$$$\left.\mathrm{1}\right){L}\left({f}^{'} \right)={x}\:{L}\left({f}\right)−{f}\left(\mathrm{0}^{+} \right) \\ $$$$\left.\mathrm{2}\right){L}\left({f}^{''} \right)={x}^{\mathrm{2}} {L}\left({f}\right)\:−{xf}\left(\mathrm{0}^{+} \right)−{f}^{'} \left(\mathrm{0}^{+} \right) \\ $$$${L}\:{means}\:{Laplace}\:{transform}. \\ $$

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