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Question Number 38154 Answers: 1 Comments: 0

Given that a,b,c are 3 consecutive term of a Geometric sequence f(n) , show that log a,logb,logc are the first 3 terms of an Arithmetic SequenceP(n).

$${Given}\:{that}\: \\ $$$$\:\:{a},{b},{c}\:{are}\:\mathrm{3}\:{consecutive}\:{term}\:{of}\: \\ $$$${a}\:{Geometric}\:{sequence}\:{f}\left({n}\right)\:,\:{show} \\ $$$${that}\:{log}\:{a},{logb},{logc}\:{are}\:{the}\:{first}\: \\ $$$$\mathrm{3}\:{terms}\:{of}\:{an}\:{Arithmetic}\:{SequenceP}\left({n}\right). \\ $$

Question Number 38151 Answers: 3 Comments: 8

a > b > 0 a^2 cos θ−b^2 sin θ=(a^2 −b^2 )sin θcos θ Find θ in terms of a, b. θ ∈ (0,(π/2)) .

$${a}\:>\:{b}\:>\:\mathrm{0} \\ $$$${a}^{\mathrm{2}} \mathrm{cos}\:\theta−{b}^{\mathrm{2}} \mathrm{sin}\:\theta=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$${Find}\:\theta\:{in}\:{terms}\:{of}\:{a},\:{b}. \\ $$$$\:\:\theta\:\in\:\left(\mathrm{0},\frac{\pi}{\mathrm{2}}\right)\:. \\ $$

Question Number 38130 Answers: 4 Comments: 0

1. ∫tan^3 (2x)sec^5 (2x) dx 2. ∫_0 ^(π/3) tan^5 (x)sec^6 (x) dx 3. ∫tan^6 (ay) dy

$$\mathrm{1}.\:\int\mathrm{tan}^{\mathrm{3}} \left(\mathrm{2}{x}\right)\mathrm{sec}^{\mathrm{5}} \left(\mathrm{2}{x}\right)\:{dx}\: \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{tan}^{\mathrm{5}} \left({x}\right)\mathrm{sec}^{\mathrm{6}} \left({x}\right)\:{dx}\: \\ $$$$\mathrm{3}.\:\int\mathrm{tan}^{\mathrm{6}} \left({ay}\right)\:{dy}\: \\ $$

Question Number 38127 Answers: 0 Comments: 1

calculate ∫_0 ^∞ e^(−x) (√(1+e^(−2x) ))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{x}} \sqrt{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx} \\ $$

Question Number 38126 Answers: 1 Comments: 2

calculate ∫_0 ^∞ e^(−2x) (√(1+e^(−4x) ))dx .

$${calculate}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\mathrm{2}{x}} \sqrt{\mathrm{1}+{e}^{−\mathrm{4}{x}} }{dx}\:. \\ $$

Question Number 38125 Answers: 1 Comments: 1

let α>0 find ∫_0 ^∞ (e^(−αx) /(√(1+e^(−2αx) )))dx .

$${let}\:\alpha>\mathrm{0}\:{find} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{e}^{−\alpha{x}} }{\sqrt{\mathrm{1}+{e}^{−\mathrm{2}\alpha{x}} }}{dx}\:. \\ $$

Question Number 38124 Answers: 1 Comments: 0

prove that ∫ (dx/(√(1+x^2 ))) =ln(x+(√(1+x^2 ))) +c 2) find ∫ (dx/(√(a+x^2 ))) with a>0

$${prove}\:{that}\:\:\int\:\:\:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+{c} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int\:\:\:\frac{{dx}}{\sqrt{{a}+{x}^{\mathrm{2}} }}\:{with}\:{a}>\mathrm{0} \\ $$

Question Number 38123 Answers: 2 Comments: 1

f is a function positive and C^1 1) find ∫ (f^′ /(2(√f)(√(1+f))))dx 2)let A_n = ∫_0 ^1 (x^(n/2) /(x(√(1+x^n )))) calculate A_n and lim_(n→+∞) A_n

$${f}\:{is}\:{a}\:{function}\:{positive}\:\:{and}\:{C}^{\mathrm{1}} \:\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\:\:\frac{{f}^{'} }{\mathrm{2}\sqrt{{f}}\sqrt{\mathrm{1}+{f}}}{dx} \\ $$$$\left.\mathrm{2}\right){let}\:\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}^{\frac{{n}}{\mathrm{2}}} }{{x}\sqrt{\mathrm{1}+{x}^{{n}} }} \\ $$$${calculate}\:{A}_{{n}} \:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$

Question Number 38122 Answers: 1 Comments: 0

calculate ∫_0 ^π ((sinx)/(√(1+cos^2 x)))dx

$${calculate}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{sinx}}{\sqrt{\mathrm{1}+{cos}^{\mathrm{2}} {x}}}{dx} \\ $$

Question Number 38121 Answers: 0 Comments: 5

let x>0 find F(x) = ∫_(−∞) ^(+∞) ((arctan(xt^2 ))/(1+t^2 ))dt

$${let}\:{x}>\mathrm{0}\:{find}\:{F}\left({x}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{arctan}\left({xt}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\: \\ $$

Question Number 38120 Answers: 0 Comments: 2

let n from N and find the value of A_n = ∫_1 ^(+∞) (dt/(t^n (√(t−1))))

$${let}\:\:{n}\:{from}\:{N}\:{and} \\ $$$${find}\:{the}\:{value}\:{of}\:\:{A}_{{n}} =\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dt}}{{t}^{{n}} \sqrt{{t}−\mathrm{1}}} \\ $$

Question Number 38119 Answers: 0 Comments: 1

calculate ∫_1 ^(+∞) (dx/(x^4 (√(x−1))))

$${calculate}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{{x}−\mathrm{1}}} \\ $$

Question Number 38118 Answers: 0 Comments: 1

prove that ∫_0 ^1 (1/(1+(t^a /2)))dt =Σ_(n=0) ^∞ (((−1)^n )/(2^n (na+1))) 2) find the value of Σ_(n=0) ^∞ (((−1)^n )/(2^n (3n+1)))

$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{{t}^{{a}} }{\mathrm{2}}}{dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} \left({na}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} \left(\mathrm{3}{n}+\mathrm{1}\right)} \\ $$

Question Number 38117 Answers: 0 Comments: 0

let x and y from R prove that ∣cos(x+iy)∣=cos^2 x +sh^2 y ∣sin(x+iy)∣^2 =sin^2 x +sh^2 y

$${let}\:{x}\:{and}\:{y}\:{from}\:{R}\:{prove}\:{that} \\ $$$$\mid{cos}\left({x}+{iy}\right)\mid={cos}^{\mathrm{2}} {x}\:+{sh}^{\mathrm{2}} {y} \\ $$$$\mid{sin}\left({x}+{iy}\right)\mid^{\mathrm{2}} ={sin}^{\mathrm{2}} {x}\:+{sh}^{\mathrm{2}} {y} \\ $$

Question Number 38116 Answers: 0 Comments: 1

find I = ∫_0 ^∞ ((cos(λx))/(ch(2x)))dx

$${find}\:\:\:\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\lambda{x}\right)}{{ch}\left(\mathrm{2}{x}\right)}{dx}\: \\ $$

Question Number 38115 Answers: 0 Comments: 0

find ∫_0 ^∞ ((sin(2x))/(sh(3x)))dx

$${find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{sh}\left(\mathrm{3}{x}\right)}{dx}\: \\ $$

Question Number 38114 Answers: 0 Comments: 2

let I_n = ∫_0 ^(2π) (dx/((p +cost)^n )) with p>1 find the value of I_n

$${let}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\left({p}\:+{cost}\right)^{{n}} }\:\:{with}\:{p}>\mathrm{1} \\ $$$${find}\:{the}\:{value}\:{of}\:{I}_{{n}} \\ $$

Question Number 38113 Answers: 0 Comments: 2

let p>1 calculate ∫_0 ^(2π) (dt/((p +cost)^2 ))

$${let}\:{p}>\mathrm{1}\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{{dt}}{\left({p}\:+{cost}\right)^{\mathrm{2}} } \\ $$

Question Number 38112 Answers: 1 Comments: 1

prove that arctan(x)= (i/2)ln(((i+x)/(i−x))) for ∣x∣<1

$${prove}\:{that}\:\:{arctan}\left({x}\right)=\:\frac{{i}}{\mathrm{2}}{ln}\left(\frac{{i}+{x}}{{i}−{x}}\right)\:{for}\:\mid{x}\mid<\mathrm{1} \\ $$

Question Number 38111 Answers: 1 Comments: 1

find lim_(x→0) ((e^x −[x])/x)

$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{e}^{{x}} \:−\left[{x}\right]}{{x}} \\ $$

Question Number 38110 Answers: 0 Comments: 0

let x from R find the value of f(x)= ∫_0 ^π ln(x^2 −2x cosθ +1)dθ

$${let}\:{x}\:{from}\:{R}\:{find}\:{the}\:{value}\:{of} \\ $$$${f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\pi} {ln}\left({x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}\right){d}\theta \\ $$

Question Number 38109 Answers: 0 Comments: 2

1) find S(x) = Σ_(n=1) ^∞ ((cos(nx))/n) 2) find Σ_(n=1) ^∞ (((−1)^n )/n)

$$\left.\mathrm{1}\right)\:{find}\:\:{S}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{cos}\left({nx}\right)}{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}} \\ $$$$ \\ $$

Question Number 38108 Answers: 0 Comments: 1

find Σ_(n=1) ^∞ (((−1)^n )/n^2 )

$${find}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$

Question Number 38107 Answers: 0 Comments: 0

find C = Σ_(n=1) ^∞ ((cos(nx))/n^2 )dx and S=Σ_(n=1) ^∞ ((sin(nx))/n^2 )

$${find}\:{C}\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }{dx}\:\:{and}\:{S}=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}^{\mathrm{2}} } \\ $$

Question Number 38106 Answers: 0 Comments: 1

calculate ∫_0 ^(+∞) e^(−3t) ln(1+e^t )dt .

$${calculate}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−\mathrm{3}{t}} {ln}\left(\mathrm{1}+{e}^{{t}} \right){dt}\:. \\ $$

Question Number 38105 Answers: 1 Comments: 0

find ∫ (dx/((√(2x+1)) +(√(2x−1))))

$${find}\:\int\:\:\:\:\:\frac{{dx}}{\sqrt{\mathrm{2}{x}+\mathrm{1}}\:+\sqrt{\mathrm{2}{x}−\mathrm{1}}}\: \\ $$

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