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Question Number 137217    Answers: 0   Comments: 0

Q137014

$${Q}\mathrm{137014} \\ $$

Question Number 137210    Answers: 1   Comments: 0

Question Number 137209    Answers: 1   Comments: 0

Question Number 137208    Answers: 0   Comments: 0

If x,y > 0 then prove that 3(√((x^2 +y^2 )/2)) + (√(xy)) ≥ 2(x+y)

$$\mathrm{If}\:\mathrm{x},\mathrm{y}\:>\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\:\mathrm{3}\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{2}}}\:+\:\sqrt{\mathrm{xy}}\:\geqslant\:\mathrm{2}\left(\mathrm{x}+\mathrm{y}\right)\: \\ $$

Question Number 137206    Answers: 2   Comments: 0

Question Number 137203    Answers: 2   Comments: 0

∫ ((ln(1+x))/x)=?

$$\int\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}=? \\ $$

Question Number 137197    Answers: 1   Comments: 0

α , β ε (0 , (π/2)) tan^2 α = 1+2tan^2 β ⇒(√2)cosα−cosβ=?

$$\alpha\:,\:\beta\:\epsilon\:\left(\mathrm{0}\:,\:\frac{\pi}{\mathrm{2}}\right) \\ $$$${tan}^{\mathrm{2}} \alpha\:=\:\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} \beta\:\:\Rightarrow\sqrt{\mathrm{2}}{cos}\alpha−{cos}\beta=? \\ $$

Question Number 137192    Answers: 2   Comments: 0

∫ ((cos x)/(1+cos x+sin x)) dx =?

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

Question Number 137191    Answers: 0   Comments: 0

Some useful approximations of sine function sin((π/7))=((96)/(221)) sin((π/9))=((128)/(373)) sin((π/(11)))=((32)/(113)) ... I am counting more ..thanking you!

$${Some}\:{useful}\:{approximations}\:{of}\:{sine}\:{function} \\ $$$${sin}\left(\frac{\pi}{\mathrm{7}}\right)=\frac{\mathrm{96}}{\mathrm{221}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{9}}\right)=\frac{\mathrm{128}}{\mathrm{373}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{11}}\right)=\frac{\mathrm{32}}{\mathrm{113}} \\ $$$$... \\ $$$${I}\:{am}\:{counting}\:{more}\:..{thanking}\:{you}! \\ $$

Question Number 137190    Answers: 1   Comments: 0

....advanced ....... calculus..... prove that::: 𝛗=∫_0 ^( 1) ln(ln((1/x))).(dx/( (√(ln((1/x)))))) =−(√π) (γ+ln(4))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{advanced}\:.......\:\:{calculus}..... \\ $$$$\:\:\:\:\:\:\:\:{prove}\:{that}::: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({ln}\left(\frac{\mathrm{1}}{{x}}\right)\right).\frac{{dx}}{\:\sqrt{{ln}\left(\frac{\mathrm{1}}{{x}}\right)}}\:=−\sqrt{\pi}\:\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$$$ \\ $$

Question Number 137189    Answers: 0   Comments: 0

Question Number 137187    Answers: 2   Comments: 0

I_n =∫_0 ^(π/2) sin^n x dx Write a relation between I_(n+2) and I_n .

$${I}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{{n}} {x}\:{dx} \\ $$$${Write}\:{a}\:{relation}\:{between}\:{I}_{{n}+\mathrm{2}} \:{and}\:{I}_{{n}} . \\ $$

Question Number 137185    Answers: 0   Comments: 0

∫_1 ^x e^((1−ln^2 t)^(1/n) ) dt=...? n an integer

$$\int_{\mathrm{1}} ^{{x}} {e}^{\left(\mathrm{1}−{ln}^{\mathrm{2}} {t}\right)^{\frac{\mathrm{1}}{{n}}} } {dt}=...?\:{n}\:{an}\:{integer} \\ $$

Question Number 137177    Answers: 0   Comments: 0

......advanced .... calculus.... Φ=Σ_(k=1) ^∞ ((ψ′(k))/k) =Σ_(n=1) ^∞ (a/n^b ) a , b =?? (adapted from brilliant) ................ ψ(k)=^(??) −γ+∫_0 ^( 1) (((1−t^(k−1) )/(1−t)))dt ∴ ψ′(k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/(1−t))dt Σ_(k=1) ^∞ ((ψ′(k))/k)=∫_0 ^( 1) ((−t^(k−1) ln(t))/((1−t)k))dt =∫_0 ^( 1) ((ln(t))/(t(1−t)))(−(t^k /k))dt=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗 𝛗=∫_0 ^( 1) ((ln(t)ln(1−t))/(t(1−t)))dt=𝛗_1 +𝛗_2 where ... ={∫_0 ^( 1) ((ln(t).ln(1−t))/(1−t))dt=𝛗_1 }+{∫_0 ^( 1) ((ln(1−t).ln(t))/t)dt=𝛗_2 } 𝛗_1 =[−(1/2)ln(t)ln^2 (1−t)]_0 ^1 +(1/2)∫_0 ^( 1) ((ln^2 (1−t))/t)dt ∴ 𝛗_1 = (1/2)(2ζ(3))=ζ(3)=^(easy) 𝛗_2 note : ∫_0 ^( 1) ((ln^2 (1−t))/t)dt=2ζ(3) (derived earlier) 𝛗=𝛗_1 +𝛗_2 =2ζ(3)=Σ_(n=1) ^∞ (2/n^3 ) .... 𝚽= Σ_(n=1) ^∞ (a/n^b ) ......⇒a=2 , b=3

$$\:\:\:\:\:\:\:\:\:\:\:......{advanced}\:\:\:\:\:....\:\:\:\:\:{calculus}.... \\ $$$$\:\:\:\:\:\:\:\Phi=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi'\left({k}\right)}{{k}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}}{{n}^{{b}} } \\ $$$$\:\:\:\:\:\:{a}\:,\:{b}\:=??\:\left({adapted}\:{from}\:{brilliant}\right) \\ $$$$\:\:\:\:\:\:\:................ \\ $$$$\:\:\:\:\:\:\:\psi\left({k}\right)\overset{??} {=}−\gamma+\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}−{t}^{{k}−\mathrm{1}} }{\mathrm{1}−{t}}\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\psi'\left({k}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{t}^{{k}−\mathrm{1}} {ln}\left({t}\right)}{\mathrm{1}−{t}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\psi'\left({k}\right)}{{k}}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{−{t}^{{k}−\mathrm{1}} {ln}\left({t}\right)}{\left(\mathrm{1}−{t}\right){k}}{dt} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right)}{{t}\left(\mathrm{1}−{t}\right)}\left(−\frac{{t}^{{k}} }{{k}}\right){dt}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right)}{{t}\left(\mathrm{1}−{t}\right)}{dt}=\boldsymbol{\phi} \\ $$$$\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right){ln}\left(\mathrm{1}−{t}\right)}{{t}\left(\mathrm{1}−{t}\right)}{dt}=\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} \:\:{where}\:... \\ $$$$\:\:\:\:\:\:\:=\left\{\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({t}\right).{ln}\left(\mathrm{1}−{t}\right)}{\mathrm{1}−{t}}{dt}=\boldsymbol{\phi}_{\mathrm{1}} \right\}+\left\{\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right).{ln}\left({t}\right)}{{t}}{dt}=\boldsymbol{\phi}_{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}_{\mathrm{1}} =\left[−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({t}\right){ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$$\:\:\:\:\:\therefore\:\:\boldsymbol{\phi}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\zeta\left(\mathrm{3}\right)\right)=\zeta\left(\mathrm{3}\right)\overset{{easy}} {=}\boldsymbol{\phi}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{note}\::\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{t}\right)}{{t}}{dt}=\mathrm{2}\zeta\left(\mathrm{3}\right)\:\left({derived}\:{earlier}\right) \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} =\mathrm{2}\zeta\left(\mathrm{3}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\:.... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Phi}=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{a}}{{n}^{{b}} }\:\:......\Rightarrow{a}=\mathrm{2}\:\:,\:{b}=\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 137173    Answers: 1   Comments: 1

Question Number 137171    Answers: 1   Comments: 0

Question Number 137157    Answers: 0   Comments: 6

Question Number 137155    Answers: 2   Comments: 0

......nice calculus ...... evaluate :: 𝛗=∫_0 ^( 2π) (1/(1+cos^4 (x)))dx=???

$$\:\:\:\:\:\:\:\:......{nice}\:\:\:\:{calculus}\:...... \\ $$$$\:\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{1}+{cos}^{\mathrm{4}} \left({x}\right)}{dx}=??? \\ $$

Question Number 137156    Answers: 0   Comments: 0

∫_1 ^∞ ((√(3x^4 +5x^3 +1))/(4x^3 +x^2 +2)) dx = ...

$$\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\sqrt{\mathrm{3}{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{3}} +\mathrm{1}}}{\mathrm{4}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{2}}\:{dx}\:=\:... \\ $$

Question Number 137148    Answers: 1   Comments: 0

Question Number 137141    Answers: 2   Comments: 0

∫_0 ^1 ((t^4 (1−t)^4 )/(1+t^2 ))dt

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{4}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$

Question Number 137139    Answers: 1   Comments: 0

Question Number 137133    Answers: 0   Comments: 0

Q137026

$${Q}\mathrm{137026} \\ $$

Question Number 137129    Answers: 1   Comments: 0

(x^2 −1)y′′−n(n+1)y=0 , n∈N Find solution that can be expanded in series help me

$$ \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{y}''−\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}=\mathrm{0}\:\:\:,\:\:\mathrm{n}\in\mathbb{N} \\ $$$$\mathrm{Find}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expanded}\:\mathrm{in}\:\mathrm{series} \\ $$$$\mathrm{help}\:\mathrm{me} \\ $$

Question Number 137123    Answers: 2   Comments: 4

∫_0 ^( 1) (x/(1+x^8 )) dx =?

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}}{\mathrm{1}+\mathrm{x}^{\mathrm{8}} }\:\mathrm{dx}\:=? \\ $$

Question Number 137117    Answers: 2   Comments: 0

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