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Question Number 128276    Answers: 1   Comments: 0

∫_e^2 ^( ∞) (dx/(x^3 ln x)) ?

$$\:\int_{{e}^{\mathrm{2}} } ^{\:\infty} \:\frac{{dx}}{{x}^{\mathrm{3}} \:\mathrm{ln}\:{x}}\:? \\ $$

Question Number 128262    Answers: 2   Comments: 0

Question Number 128251    Answers: 1   Comments: 0

Question Number 128244    Answers: 0   Comments: 0

...nice calculus... prove that ::Ω= ∫_0 ^(π/4) ln(sin(x))d=((−π)/4)log(2)−(G/2) log(2sin(x))=Σ_(n=1) ^∞ ((−1)/n)cos(2nx) Ω= ∫_0 ^( (π/4)) {−log(2)−Σ_(n=1) ^∞ ((cos(2nx))/n)}dx =((−π)/4)log(2)−Σ_(n=1) ^∞ ∫_0 ^( (π/4)) ((cos(2nx))/n)dx =((−π)/4)log(2)−Σ_(n=1) ^∞ [(1/(2n^2 ))sin(2nx)]_0 ^(π/4) =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ ((sin(((nπ)/2)))/n^2 ) =((−π)/4)log(2)−(1/2){(1/1^2 )−(1/3^2 ) +(1/5^2 )−..} =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 )) ∴ Ω =((−π)/4)log(2)−(G/2) ✓ G:= catalan constant...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:\:::\Omega=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({x}\right)\right){d}=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}} \\ $$$$\:\:\:\:{log}\left(\mathrm{2}{sin}\left({x}\right)\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{1}}{{n}}{cos}\left(\mathrm{2}{nx}\right) \\ $$$$\:\Omega=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left\{−{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\right\}{dx} \\ $$$$=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}{dx} \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }{sin}\left(\mathrm{2}{nx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\:\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} } \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−..\right\} \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\Omega\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}}\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{G}:=\:{catalan}\:\:{constant}... \\ $$

Question Number 128221    Answers: 1   Comments: 0

Given f(x+(1/x)) = x^4 −(1/x^4 )+2 then ∫_1 ^( 2) (1−x^(−2) )f(x)dx=

$$\:{Given}\:{f}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:=\:{x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2} \\ $$$$\:{then}\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{−\mathrm{2}} \right){f}\left({x}\right){dx}= \\ $$

Question Number 128194    Answers: 1   Comments: 2

Question Number 128192    Answers: 0   Comments: 0

Question Number 128175    Answers: 1   Comments: 1

∫_0 ^( 1) ∫_0 ^( 1) ((dx dy)/(1−xy^3 )) ?

$$\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{dx}\:{dy}}{\mathrm{1}−{xy}^{\mathrm{3}} }\:? \\ $$

Question Number 128109    Answers: 1   Comments: 0

∅ = ∫ (x−2) (√((x+1)/(x−1))) dx

$$\emptyset\:=\:\int\:\left(\mathrm{x}−\mathrm{2}\right)\:\sqrt{\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}}\:\mathrm{dx}\: \\ $$

Question Number 128082    Answers: 0   Comments: 0

β = ∫ ((1+ln (x))/(x.cos^2 (x))) dx

$$\beta\:=\:\int\:\frac{\mathrm{1}+\mathrm{ln}\:\left(\mathrm{x}\right)}{\mathrm{x}.\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$

Question Number 128079    Answers: 1   Comments: 0

Ω = ∫ ln (x+(√(x^2 +a^2 )) )dx

$$\Omega\:=\:\int\:\mathrm{ln}\:\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\:\right)\mathrm{dx} \\ $$

Question Number 128073    Answers: 1   Comments: 0

∫ ((sin(2x))/((1 − x)^3 )) dx

$$\int\:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\left(\mathrm{1}\:\:−\:\:\mathrm{x}\right)^{\mathrm{3}} }\:\:\mathrm{dx} \\ $$

Question Number 128057    Answers: 3   Comments: 0

∫_0 ^( 1) ∫_0 ^( 1) (1/(1−xy^2 )) dx dy =?

$$\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{xy}^{\mathrm{2}} }\:\mathrm{dx}\:\mathrm{dy}\:=? \\ $$

Question Number 128052    Answers: 3   Comments: 0

I = ∫ arctan (((x−2)/(x+2))) dx

$$\:\mathrm{I}\:=\:\int\:\mathrm{arctan}\:\left(\frac{\mathrm{x}−\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)\:\mathrm{dx}\: \\ $$

Question Number 128026    Answers: 0   Comments: 0

∫_(π/2) ^((2π)/3) (((arcsinhx)^2 )/(2cosx))dx

$$\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{2}\pi}{\mathrm{3}}} \frac{\left(\mathrm{arcsinhx}\right)^{\mathrm{2}} }{\mathrm{2cosx}}\mathrm{dx} \\ $$

Question Number 128025    Answers: 2   Comments: 0

Question Number 128023    Answers: 0   Comments: 0

....nice calculus...= Titu′s lemma:: for any positive numbers : a_1 ,a_2 ,...,a_n , b_1 ,b_2 ,...,b_n we have: (((a_1 +...+a_n )^2 )/(b_1 +...+b_n ))≤(a_1 ^2 /b_1 ) +...+(a_n ^2 /b_n ) proof : put : x=(x_1 ,...,x_n )∈R^n :y=(y_1 ,...,y_n )∈R^n (x.y)^2 ≤∣x∣^2 ∣y∣^2 (cauchy−schwarz inequality) (x_1 y_1 +...+x_n y_n )^2 ≤(x_(1 ) ^2 +...+x_n ^2 )(y_1 ^2 +...+y_(n ) ^2 ) by applying subsitution : x_i =(a_i /( (√b_i ))) , y_i =(√b_i ) (i=1,2 ,...,n) ((a_(1 ) ^2 +...+a_(n ) ^2 )/(b_2 +...+b_n ))≤(a_1 ^2 /b_1 )+...+(a_n ^2 /b_n ) ✓✓

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}...= \\ $$$$\:\:{Titu}'{s}\:{lemma}:: \\ $$$$\:{for}\:{any}\:{positive}\:{numbers}\:: \\ $$$${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,...,{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,...,{b}_{{n}} \\ $$$$\:{we}\:{have}: \\ $$$$\:\frac{\left({a}_{\mathrm{1}} +...+{a}_{{n}} \right)^{\mathrm{2}} }{{b}_{\mathrm{1}} +...+{b}_{{n}} }\leqslant\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{{b}_{\mathrm{1}} }\:+...+\frac{{a}_{{n}} ^{\mathrm{2}} }{{b}_{{n}} }\: \\ $$$${proof}\:: \\ $$$${put}\::\:{x}=\left({x}_{\mathrm{1}} ,...,{x}_{{n}} \right)\in\mathbb{R}^{{n}} \\ $$$$\:\:\:\:\:\:\:\:\::{y}=\left({y}_{\mathrm{1}} ,...,{y}_{{n}} \right)\in\mathbb{R}^{{n}} \\ $$$$\left({x}.{y}\right)^{\mathrm{2}} \leqslant\mid{x}\mid^{\mathrm{2}} \mid{y}\mid^{\mathrm{2}} \left({cauchy}−{schwarz}\:\:{inequality}\right) \\ $$$$\left({x}_{\mathrm{1}} {y}_{\mathrm{1}} +...+{x}_{{n}} {y}_{{n}} \right)^{\mathrm{2}} \leqslant\left({x}_{\mathrm{1}\:\:\:} ^{\mathrm{2}} +...+{x}_{{n}} ^{\mathrm{2}} \right)\left({y}_{\mathrm{1}} ^{\mathrm{2}} +...+{y}_{{n}\:} ^{\mathrm{2}} \right) \\ $$$$\:{by}\:{applying}\:{subsitution}\:: \\ $$$$\:{x}_{{i}} =\frac{{a}_{{i}} }{\:\sqrt{{b}_{{i}} }}\:\:\:,\:\:{y}_{{i}} =\sqrt{{b}_{{i}} }\:\left({i}=\mathrm{1},\mathrm{2}\:,...,{n}\right) \\ $$$$\:\frac{{a}_{\mathrm{1}\:} ^{\mathrm{2}} +...+{a}_{{n}\:} ^{\mathrm{2}} }{{b}_{\mathrm{2}} +...+{b}_{{n}} }\leqslant\frac{{a}_{\mathrm{1}} ^{\mathrm{2}} }{{b}_{\mathrm{1}} }+...+\frac{{a}_{{n}} ^{\mathrm{2}} }{{b}_{{n}} }\:\:\:\checkmark\checkmark \\ $$$$ \\ $$

Question Number 127958    Answers: 1   Comments: 0

...nice calculus... calculate Ω=∫_1 ^( ∞) ((ln(x^4 −2x^2 +2))/(x(√(x^2 −1)) )) dx=?

$$\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{calculate} \\ $$$$\:\:\:\:\Omega=\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}\right)}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:}\:{dx}=? \\ $$$$ \\ $$

Question Number 127948    Answers: 1   Comments: 0

Question Number 127952    Answers: 1   Comments: 0

Question Number 127925    Answers: 1   Comments: 0

find F(a)=∫_0 ^1 (√((1+a^2 t^2 )/(1−t^2 ))) dt for background see Q127811.

$${find}\:{F}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt} \\ $$$$ \\ $$$${for}\:{background}\:{see}\:{Q}\mathrm{127811}. \\ $$

Question Number 127904    Answers: 2   Comments: 0

prove that ∫_0 ^( 100) (dx/( (√(x(100−x))))) = π

$$\:\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\:\mathrm{100}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}\left(\mathrm{100}−\mathrm{x}\right)}}\:=\:\pi \\ $$

Question Number 127885    Answers: 1   Comments: 0

∫(((sin (2tan^(−1) (x)+x))/x)) the limit [0,∞)

$$\int\left(\frac{\mathrm{sin}\:\left(\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)+{x}\right)}{{x}}\right)\:\:{the}\:{limit}\:\left[\mathrm{0},\infty\right) \\ $$

Question Number 127870    Answers: 2   Comments: 0

2021 HAPPY NEW Year 1)∫((x^3 +3x+2)/((x^2 +1)^2 (x+1)))dx 2)∫((2cos(x)−sin(x))/(3sin(x)+5cos(x)))dx 3)∫((tan(2x))/( (√(sin^6 (x)+cos^6 (x)))))dx 4)∫x(√((1−x^2 )/(1+x^2 ))) dx

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2021} \\ $$$${HAPPY}\:{NEW}\:{Year} \\ $$$$\left.\mathrm{1}\right)\int\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{1}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\int\frac{\mathrm{2}{cos}\left({x}\right)−{sin}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{5}{cos}\left({x}\right)}{dx} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\int\frac{{tan}\left(\mathrm{2}{x}\right)}{\:\sqrt{{sin}^{\mathrm{6}} \left({x}\right)+{cos}^{\mathrm{6}} \left({x}\right)}}{dx} \\ $$$$ \\ $$$$\left.\mathrm{4}\right)\int{x}\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$

Question Number 127857    Answers: 0   Comments: 1

∫(√x)e^x dx ?

$$\int\sqrt{{x}}{e}^{{x}} {dx}\:\:? \\ $$

Question Number 127851    Answers: 1   Comments: 0

ψ = ∫ (dx/(x^3 (((x^5 +1)^3 ))^(1/5) )) ?

$$\:\psi\:=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \:\sqrt[{\mathrm{5}}]{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{1}\right)^{\mathrm{3}} }}\:?\: \\ $$

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