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Question Number 85696    Answers: 0   Comments: 0

Montrer que: (√5)+(√(30))+(√(50))<(√(10))+(√(20))+(√(60)) {niveau second)

$$\mathrm{Montrer}\:\mathrm{que}: \\ $$$$\sqrt{\mathrm{5}}+\sqrt{\mathrm{30}}+\sqrt{\mathrm{50}}<\sqrt{\mathrm{10}}+\sqrt{\mathrm{20}}+\sqrt{\mathrm{60}} \\ $$$$\left\{\mathrm{niveau}\:\mathrm{second}\right) \\ $$

Question Number 85695    Answers: 0   Comments: 0

Question Number 85694    Answers: 0   Comments: 1

log_(((x/(x−3)))) (7) < log_(((x/3))) (7)

$$\mathrm{log}_{\left(\frac{{x}}{{x}−\mathrm{3}}\right)} \left(\mathrm{7}\right)\:<\:\mathrm{log}_{\left(\frac{{x}}{\mathrm{3}}\right)} \:\left(\mathrm{7}\right)\: \\ $$

Question Number 85677    Answers: 2   Comments: 3

∫_0 ^π ((sin (((21x)/2)))/(sin ((x/2)))) dx

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\frac{\mathrm{sin}\:\left(\frac{\mathrm{21}{x}}{\mathrm{2}}\right)}{\mathrm{sin}\:\left(\frac{{x}}{\mathrm{2}}\right)}\:{dx}\: \\ $$

Question Number 85676    Answers: 0   Comments: 15

∫ _0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) let x = tan t ⇒dx=sec^2 t dt ∫_0 ^(π/2) ((sec^2 t dt)/((tan t+sec t)^2 )) = ∫_0 ^(π/2) (dt/((sin t+1)^2 )) = ∫_0 ^(π/2) (dt/((cos (1/2)t+sin (1/2)t)^4 )) = ∫_0 ^(π/2) (dt/(4cos^4 ((1/2)t−(π/4)))) = (1/4)∫_0 ^(π/2) sec^4 ((1/2)t−(π/4)) dt [ let (1/2)t−(π/4)= u] = (1/4)∫_(−(π/4)) ^0 sec^4 u ×2du =(1/2)∫ _(−(π/4)) ^0 (tan^2 u+1) d(tan u) = (1/2) [(1/3)tan^3 u + tan u ]_(−(π/4)) ^0 = (1/2) [ 0−(−(1/3)−1)]= (2/3)

$$\int\underset{\mathrm{0}} {\overset{\infty} {\:}}\:\frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$${let}\:{x}\:=\:\mathrm{tan}\:{t}\:\Rightarrow{dx}=\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {t}\:{dt}}{\left(\mathrm{tan}\:{t}+\mathrm{sec}\:{t}\right)^{\mathrm{2}} }\:=\: \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{sin}\:{t}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\left(\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{t}+\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{t}\right)^{\mathrm{4}} } \\ $$$$=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dt}}{\mathrm{4cos}^{\mathrm{4}} \:\left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}\right)\:{dt} \\ $$$$\left[\:{let}\:\frac{\mathrm{1}}{\mathrm{2}}{t}−\frac{\pi}{\mathrm{4}}=\:{u}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\int}}\:\mathrm{sec}\:^{\mathrm{4}} {u}\:×\mathrm{2}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\mathrm{0}} {\:}}\left(\mathrm{tan}\:^{\mathrm{2}} {u}+\mathrm{1}\right)\:{d}\left(\mathrm{tan}\:{u}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} {u}\:+\:\mathrm{tan}\:{u}\:\right]_{−\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\left[\:\mathrm{0}−\left(−\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}\right)\right]=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Question Number 85670    Answers: 0   Comments: 3

Question Number 85669    Answers: 1   Comments: 4

∫ ((√(1+x))/(√(1−x))) dx

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

Question Number 85668    Answers: 0   Comments: 2

z = 2 + i find arg(z)

$$\mathrm{z}\:=\:\mathrm{2}\:+\:\mathrm{i}\: \\ $$$$\mathrm{find}\:\mathrm{arg}\left(\mathrm{z}\right) \\ $$

Question Number 85667    Answers: 1   Comments: 0

∫ (dx/(√(1−sin 2x)))

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

Question Number 85666    Answers: 0   Comments: 0

Question Number 85664    Answers: 0   Comments: 2

Show that a group order 100 is not simple

$$\boldsymbol{{S}\mathrm{how}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{group}}\:\boldsymbol{\mathrm{order}}\:\mathrm{100}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{simple}} \\ $$

Question Number 85656    Answers: 0   Comments: 1

Question Number 85655    Answers: 0   Comments: 0

Question Number 85653    Answers: 0   Comments: 1

if f≥0 and (d/dx)(f(x))^2 =(f′(x))^2 and f(0)=1 find f(x) if f(x)=((4x^3 )/(x^2 +1)) find f^( −1) (x) and (f^( −1) )^′ (2)

$${if}\:\:{f}\geqslant\mathrm{0}\:\:{and}\:\:\:\frac{{d}}{{dx}}\left({f}\left({x}\right)\right)^{\mathrm{2}} =\left({f}'\left({x}\right)\right)^{\mathrm{2}} \:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${find}\:{f}\left({x}\right)\:\:\: \\ $$$$ \\ $$$${if}\:{f}\left({x}\right)=\frac{\mathrm{4}{x}^{\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{1}}\:\:{find}\:{f}^{\:−\mathrm{1}} \left({x}\right)\:\:\:{and}\:\left({f}^{\:−\mathrm{1}} \right)^{'} \left(\mathrm{2}\right) \\ $$

Question Number 85649    Answers: 1   Comments: 1

Proove that : ((√(2−(√3)))/2) = (((√6)−(√2))/4)

$$\mathrm{Proove}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

Question Number 85648    Answers: 2   Comments: 0

∫(((x^3 −4))/((x+1)))dx

$$\int\frac{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{4}\right)}{\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$

Question Number 85646    Answers: 0   Comments: 0

show that ∫(1/([x(x−1)(x−2)(x−3)...(x−m)]^2 ))dx= =(1/((m!)^2 ))Σ_(n=0) ^m ( ((m),(n) )^2 /(n−x))+(2/((m!)^2 ))ln∣Π_(n=0) ^m (x−n)^( ((m),(n) )^2 (H_(m−n) −H_n )) ∣+c

$${show}\:{that} \\ $$$$\int\frac{\mathrm{1}}{\left[{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)...\left({x}−{m}\right)\right]^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\left({m}!\right)^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} }{{n}−{x}}+\frac{\mathrm{2}}{\left({m}!\right)^{\mathrm{2}} }{ln}\mid\underset{{n}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}−{n}\right)^{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \left({H}_{{m}−{n}} −{H}_{{n}} \right)} \mid+{c} \\ $$

Question Number 85641    Answers: 0   Comments: 2

calculate A_λ =∫_3 ^∞ (dx/((x+λ)^3 (x−2)^4 )) (λ>0)

$${calculate}\:{A}_{\lambda} =\int_{\mathrm{3}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\lambda\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{4}} }\:\:\:\left(\lambda>\mathrm{0}\right) \\ $$

Question Number 85637    Answers: 1   Comments: 2

∫ (dx/(x+(√(x^2 +1))))

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

Question Number 85630    Answers: 0   Comments: 1

∫_0 ^(2Π) (dx/((√2)−cosx))

$$\int_{\mathrm{0}} ^{\mathrm{2}\Pi} \:\:\:\frac{{dx}}{\sqrt{\mathrm{2}}−{cosx}}\:\: \\ $$

Question Number 85625    Answers: 1   Comments: 0

lim_(x→0) ((x^n −(sin x)^n )/((sin x)^(n+2) ))

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{n}} −\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{n}} }{\left(\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{n}+\mathrm{2}} } \\ $$

Question Number 85624    Answers: 0   Comments: 0

Question Number 85623    Answers: 1   Comments: 0

prove that sin x cos 2x = (1/(4sin 4x sec x))

$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{2x}\:=\:\frac{\mathrm{1}}{\mathrm{4sin}\:\mathrm{4x}\:\mathrm{sec}\:\mathrm{x}} \\ $$

Question Number 85620    Answers: 1   Comments: 0

prove that cosh (x−y)=cosh xcosh y−sinh xsinh y

$${prove}\:{that} \\ $$$$ \\ $$$$\mathrm{cosh}\:\left({x}−{y}\right)=\mathrm{cosh}\:{x}\mathrm{cosh}\:{y}−\mathrm{sinh}\:{x}\mathrm{sinh}\:{y} \\ $$

Question Number 85606    Answers: 0   Comments: 7

Question Number 85603    Answers: 0   Comments: 0

prove the relation ∫_0 ^1 ((li_5 ((x)^(1/5) ))/(x)^(1/5) )dx=(5/4)(((25)/(3072))−((ζ(2))/2^6 )+((ζ(3))/2^4 )−((ζ(4))/2^2 )+ζ(5))

$${prove}\:{the}\:{relation} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{li}_{\mathrm{5}} \left(\sqrt[{\mathrm{5}}]{{x}}\right)}{\sqrt[{\mathrm{5}}]{{x}}}{dx}=\frac{\mathrm{5}}{\mathrm{4}}\left(\frac{\mathrm{25}}{\mathrm{3072}}−\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}^{\mathrm{6}} }+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}^{\mathrm{4}} }−\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{2}^{\mathrm{2}} }+\zeta\left(\mathrm{5}\right)\right) \\ $$

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