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Question Number 134454    Answers: 0   Comments: 0

...nice calculus... find the value of: 𝛗=Ξ£_(n=1) ^∞ ((ΞΆ(2n)βˆ’1)/(n+1))=?

$$\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}\right)βˆ’\mathrm{1}}{{n}+\mathrm{1}}=? \\ $$$$ \\ $$

Question Number 134420    Answers: 1   Comments: 0

∫^( Ο€/2) _(βˆ’Ο€/2) (1/(2019^x +1)). ((sin^(2020) x)/(sin^(2020) x+cos^(2020) x)) dx ?

$$\underset{βˆ’\pi/\mathrm{2}} {\int}^{\:\:\:\:\:\:\pi/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2019}^{{x}} +\mathrm{1}}.\:\frac{\mathrm{sin}\:^{\mathrm{2020}} {x}}{\mathrm{sin}\:^{\mathrm{2020}} {x}+\mathrm{cos}\:^{\mathrm{2020}} {x}}\:{dx}\:? \\ $$

Question Number 134418    Answers: 1   Comments: 0

𝛗=∫_0 ^( ∞) (arctan((1/x)))^2 =???

$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)^{\mathrm{2}} =??? \\ $$

Question Number 134303    Answers: 2   Comments: 0

F=∫_0 ^∞ ((16 arctan (x))/(1+x^2 )) dx

$$\mathcal{F}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{16}\:\mathrm{arctan}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$

Question Number 134302    Answers: 2   Comments: 0

∫_0 ^( 1) ((ln x)/( (√(1βˆ’x^2 )))) dx

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:{x}}{\:\sqrt{\mathrm{1}βˆ’{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$

Question Number 134301    Answers: 3   Comments: 0

Ω = ∫_0 ^∞ (x^2 /((1+x^2 )^4 )) dx

$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:{dx} \\ $$

Question Number 134291    Answers: 1   Comments: 0

Prove ∫_0 ^( ∞) (x^a /(1+e^x ))dx = (1βˆ’2^(βˆ’a) )𝛇(a+1)πšͺ(a+1)

$$ \\ $$$$\:\boldsymbol{\mathrm{Prove}}\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\mathrm{dx}\:=\:\left(\mathrm{1}βˆ’\mathrm{2}^{βˆ’\mathrm{a}} \right)\boldsymbol{\zeta}\left(\mathrm{a}+\mathrm{1}\right)\boldsymbol{\Gamma}\left(\mathrm{a}+\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 134289    Answers: 3   Comments: 0

I=∫ (x^n /( (√(ax+b)))) dx H=∫ (x^4 /( (√(2x+1)))) dx

$$\mathrm{I}=\int\:\frac{\mathrm{x}^{\mathrm{n}} }{\:\sqrt{\mathrm{ax}+\mathrm{b}}}\:\mathrm{dx} \\ $$$$\mathrm{H}=\int\:\frac{\mathrm{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{2x}+\mathrm{1}}}\:\mathrm{dx} \\ $$

Question Number 134272    Answers: 2   Comments: 0

nice calculus prove that: : i :: =∫_0 ^( ∞) ((cos(Ο€x))/(e^(2Ο€(√x) ) βˆ’1)) dx=((2βˆ’(√2) )/8) ii:: compute: Ξ£_(n=βˆ’βˆž) ^∞ (1/(n^4 +9n^2 +10)) =? ...m.n...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nice}\:\:\:\mathrm{calculus} \\ $$$$\:\:\:\:\mathrm{prove}\:\mathrm{that}:\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}\:::\:\: =\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\pi{x}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} βˆ’\mathrm{1}}\:{dx}=\frac{\mathrm{2}βˆ’\sqrt{\mathrm{2}}\:}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{compute}:\:\:\underset{{n}=βˆ’\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}... \\ $$

Question Number 134239    Answers: 1   Comments: 2

can i ask for some help? how to prove this? (1/2)<∫_0 ^(1/2) (dx/( (√(1βˆ’x^3 ))))<(Ο€/6)

$${can}\:{i}\:{ask}\:{for}\:{some}\:{help}? \\ $$$${how}\:{to}\:{prove}\:{this}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}<\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{1}βˆ’{x}^{\mathrm{3}} }}<\frac{\pi}{\mathrm{6}} \\ $$

Question Number 134227    Answers: 0   Comments: 1

I struck upon this Ξ£_(n=0) ^∞ 16^n =((βˆ’1)/(15)) ∫_0 ^( Ο€) cos^4 x sin x dx=((βˆ’1)/5)[cos^5 x]_0 ^Ο€ =0.4 in another way I=∫^ cos^4 x sin x dx I^ =cos^4 (x) (βˆ’cos x)βˆ’βˆ«4cos^3 x(βˆ’sin x)(βˆ’cos x)dx I^ =cos^5 xβˆ’4∫cos^4 x sin xdx I^ =cos^5 xβˆ’4I I=βˆ’cos^5 xΞ£_(n=0) ^∞ (βˆ’4)^n +C [I]_0 ^Ο€ =[(βˆ’(βˆ’1)Ξ£_(n=0) ^∞ (βˆ’4)^n )βˆ’(βˆ’Ξ£_(n=0) ^∞ (βˆ’4)^n )] =2Ξ£_(n=0) ^∞ (βˆ’4)^n =2(Ξ£_(n=0) ^∞ 16^n βˆ’4Ξ£_(n=0) ^∞ 16^n ) =2(βˆ’3Ξ£_(n=0) ^∞ 16^n )=βˆ’6Ξ£_(n=0) ^∞ 16^n βˆ’6Ξ£_(n=0) ^∞ 16^n =0.4β‡’Ξ£_(n=0) ^∞ 16^n =((βˆ’0.4)/6)=((βˆ’1)/(15)) did I do something wrong?

$$\mathrm{I}\:\:\mathrm{struck}\:\mathrm{upon}\:\mathrm{this} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{βˆ’\mathrm{1}}{\mathrm{15}} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=\frac{βˆ’\mathrm{1}}{\mathrm{5}}\left[\mathrm{cos}^{\mathrm{5}} \:{x}\right]_{\mathrm{0}} ^{\pi} =\mathrm{0}.\mathrm{4} \\ $$$$\mathrm{in}\:\mathrm{another}\:\mathrm{way} \\ $$$${I}=\int^{\:} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{4}} \:\left({x}\right)\:\left(βˆ’\mathrm{cos}\:{x}\right)βˆ’\int\mathrm{4cos}^{\mathrm{3}} \:{x}\left(βˆ’\mathrm{sin}\:{x}\right)\left(βˆ’\mathrm{cos}\:{x}\right){dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}βˆ’\mathrm{4}\int\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{xdx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}βˆ’\mathrm{4}{I} \\ $$$${I}=βˆ’\mathrm{cos}^{\mathrm{5}} \:{x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{4}\right)^{{n}} +{C} \\ $$$$\left[{I}\right]_{\mathrm{0}} ^{\pi} =\left[\left(βˆ’\left(βˆ’\mathrm{1}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{4}\right)^{{n}} \right)βˆ’\left(βˆ’\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{4}\right)^{{n}} \right)\right] \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{4}\right)^{{n}} =\mathrm{2}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} βˆ’\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right) \\ $$$$=\mathrm{2}\left(βˆ’\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right)=βˆ’\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \\ $$$$βˆ’\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\mathrm{0}.\mathrm{4}\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{βˆ’\mathrm{0}.\mathrm{4}}{\mathrm{6}}=\frac{βˆ’\mathrm{1}}{\mathrm{15}} \\ $$$$\mathrm{did}\:\mathrm{I}\:\mathrm{do}\:\mathrm{something}\:\mathrm{wrong}? \\ $$

Question Number 134219    Answers: 1   Comments: 0

∫_0 ^( Ο€) cos^4 x sin x dx=?

$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=? \\ $$

Question Number 134185    Answers: 1   Comments: 0

.....advanced calculus.... prove that:: i:𝛗=∫_0 ^( (Ο€/2)) ((cos(tan(x)βˆ’x))/(cos(x)))dx=(Ο€/e) ii:Ξ _(n=2) ^∞ e(1βˆ’(1/n^2 ))^n^2 =(Ο€/(e(√e)))

$$\:\:\:\:\:\:\:\:\:\:\:\:.....{advanced}\:\:\:{calculus}.... \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:{i}:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({tan}\left({x}\right)βˆ’{x}\right)}{{cos}\left({x}\right)}{dx}=\frac{\pi}{{e}} \\ $$$$\:\:\:\:{ii}:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}{e}\left(\mathrm{1}βˆ’\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } =\frac{\pi}{{e}\sqrt{{e}}} \\ $$$$\:\: \\ $$

Question Number 134182    Answers: 1   Comments: 0

Question Number 134058    Answers: 1   Comments: 0

J = ∫ (dx/(1+tan x+csc x+cot x+sec x))

$$\mathcal{J}\:=\:\int\:\frac{{dx}}{\mathrm{1}+\mathrm{tan}\:{x}+\mathrm{csc}\:{x}+\mathrm{cot}\:{x}+\mathrm{sec}\:{x}} \\ $$

Question Number 134039    Answers: 0   Comments: 0

Question Number 134038    Answers: 1   Comments: 0

Question Number 134016    Answers: 1   Comments: 0

?prove :Ξ£_(n=1) ^∞ (((βˆ’1)^(nβˆ’1) H_(2n) )/(2n+1))=(Ο€/8)ln(2)..

$$ \\ $$$$\:\:\:\:\:\:?{prove}\::\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}βˆ’\mathrm{1}} {H}_{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right).. \\ $$

Question Number 133973    Answers: 1   Comments: 0

Given { ((f(x)=((x+(√(x^2 +(1/(27))))))^(1/3) +((xβˆ’(√(x^2 +(1/(27))))))^(1/3) )),((g(x)=x^3 +x+1)) :} Find ∫_0 ^4 (gβ—‹fβ—‹g)(x) dx .

$$\:\mathrm{Given}\:\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}}+\sqrt[{\mathrm{3}}]{\mathrm{x}βˆ’\sqrt{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}}}\\{\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}}\end{cases} \\ $$$$\mathrm{Find}\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\left(\mathrm{g}\circ\mathrm{f}\circ\mathrm{g}\right)\left(\mathrm{x}\right)\:\mathrm{dx}\:. \\ $$

Question Number 133972    Answers: 1   Comments: 0

H = ∫ (((2xβˆ’1)^7 )/((2x+1)^9 )) dx

$$\mathscr{H}\:=\:\int\:\frac{\left(\mathrm{2x}βˆ’\mathrm{1}\right)^{\mathrm{7}} }{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{9}} }\:\mathrm{dx}\: \\ $$

Question Number 133963    Answers: 1   Comments: 0

Y = ∫ (dx/( ((1+x^6 ))^(1/6) ))?

$$\mathcal{Y}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{6}}]{\mathrm{1}+{x}^{\mathrm{6}} }}? \\ $$

Question Number 133957    Answers: 2   Comments: 0

1)decompose F(x)=(1/((x+1)^5 (2xβˆ’1)^4 )) 2) find ∫_1 ^∞ F(x)dx

$$\left.\mathrm{1}\right)\mathrm{decompose}\:\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{2x}βˆ’\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\int_{\mathrm{1}} ^{\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 133943    Answers: 0   Comments: 0

.......advanced calculus...... prove that: 𝛗= ∫_0 ^( ∞) ((cos(x^2 )βˆ’cos(x))/x)dx=(Ξ³/2) Ξ³: eulerβˆ’mascheroni constant...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{advanced}\:\:\:\:\:{calculus}...... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({x}^{\mathrm{2}} \right)βˆ’{cos}\left({x}\right)}{{x}}{dx}=\frac{\gamma}{\mathrm{2}} \\ $$$$\:\:\:\gamma:\:{euler}βˆ’{mascheroni}\:{constant}... \\ $$

Question Number 133911    Answers: 1   Comments: 0

A=∫ ((cos x)/(sin 5x+sin x)) dx ?

$$\mathcal{A}=\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{5x}+\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Question Number 133910    Answers: 0   Comments: 0

De^ montrer que; Ξ£_(n=1) ^∞ (((βˆ’1)^n )/(1+n^4 ))=(1/2)[(Ο€/( (√2))) ((sin((Ο€/( (√2))))cosh((Ο€/( (√2))))+sinh((Ο€/( (√2))))cos((Ο€/( (√2)))))/(sinh^2 ((Ο€/( (√2))))+sin^2 ((Ο€/( (√2))))))]

$$\:\:\:\:\mathcal{D}\acute {\mathrm{e}montrer}\:\mathrm{que}; \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{1}+\mathrm{n}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cosh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sinh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{sinh}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}\right] \\ $$

Question Number 133907    Answers: 0   Comments: 0

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