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Question Number 120064    Answers: 1   Comments: 0

I = ∫ _0 ^(π/2) ((1/(ln (tan r))) + (1/(1−tan r)) ) dr

$${I}\:=\:\int\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\:}}\left(\frac{\mathrm{1}}{\mathrm{ln}\:\left(\mathrm{tan}\:{r}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{tan}\:{r}}\:\right)\:{dr} \\ $$

Question Number 120060    Answers: 2   Comments: 0

(i) ∫_(−2) ^0 (dx/(2x+3)) (ii)∫_3 ^5 (dx/( (((4−x)^2 ))^(1/3) ))

$$\left({i}\right)\:\underset{−\mathrm{2}} {\overset{\mathrm{0}} {\int}}\:\frac{{dx}}{\mathrm{2}{x}+\mathrm{3}} \\ $$$$\left({ii}\right)\underset{\mathrm{3}} {\overset{\mathrm{5}} {\int}}\:\frac{{dx}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }}\: \\ $$

Question Number 120059    Answers: 1   Comments: 0

Question Number 120040    Answers: 2   Comments: 0

∫ (t^5 /( (√(2+t^2 )))) dt

$$\:\int\:\frac{{t}^{\mathrm{5}} }{\:\sqrt{\mathrm{2}+{t}^{\mathrm{2}} }}\:{dt}\: \\ $$

Question Number 120025    Answers: 2   Comments: 0

calculate f^′ (x) 1) f(x) =∫_0 ^∞ ((cos(xt))/(t^2 +x^2 ))dt 2)f(x)=∫_0 ^∞ ((sin(xt^2 +(√2)))/(t^2 +x^2 +3))dt

$$\mathrm{calculate}\:\mathrm{f}^{'} \left(\mathrm{x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{xt}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dt} \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\mathrm{xt}^{\mathrm{2}} +\sqrt{\mathrm{2}}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}}\mathrm{dt} \\ $$

Question Number 119989    Answers: 2   Comments: 0

Question Number 119970    Answers: 3   Comments: 0

∫ (dx/(x^2 (√(25−x^2 )))) ?

$$\:\int\:\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{25}−{x}^{\mathrm{2}} }}\:? \\ $$

Question Number 119960    Answers: 1   Comments: 0

Question Number 119934    Answers: 2   Comments: 0

∫ ((sin^8 x−cos^8 x)/(1−(1/2)sin^2 2x)) dx

$$\:\:\:\int\:\frac{\mathrm{sin}\:^{\mathrm{8}} {x}−\mathrm{cos}\:^{\mathrm{8}} {x}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:{dx}\: \\ $$

Question Number 119930    Answers: 1   Comments: 0

Question Number 119920    Answers: 5   Comments: 0

∫ (√((1−x)/(1+x))) dx = ? , xε(−1,1)

$$\:\int\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:=\:?\:,\:{x}\epsilon\left(−\mathrm{1},\mathrm{1}\right) \\ $$

Question Number 119852    Answers: 1   Comments: 0

lim_(n→∞) n^2 ∫ _0 ^(1/n) x^(x+1) dx =?

$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{\mathrm{2}} \:\int\:\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{{n}}} {\:}}{x}^{{x}+\mathrm{1}} \:{dx}\:=? \\ $$

Question Number 119784    Answers: 2   Comments: 0

∫ (dx/( (√((4x−x^2 )^3 ))))

$$\:\:\int\:\frac{{dx}}{\:\sqrt{\left(\mathrm{4}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} }} \\ $$$$ \\ $$

Question Number 119773    Answers: 3   Comments: 0

∫_(−4) ^4 x^3 (√(16−x^2 )) sec x dx

$$\underset{−\mathrm{4}} {\overset{\mathrm{4}} {\int}}\:{x}^{\mathrm{3}} \sqrt{\mathrm{16}−{x}^{\mathrm{2}} \:}\:\mathrm{sec}\:{x}\:{dx}\: \\ $$

Question Number 119821    Answers: 3   Comments: 0

∫_(−3) ^0 ((6x−6)/( (√(x^2 −2x+1)))) dx =?

$$\:\underset{−\mathrm{3}} {\overset{\mathrm{0}} {\int}}\:\frac{\mathrm{6}{x}−\mathrm{6}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}}\:{dx}\:=? \\ $$

Question Number 119762    Answers: 3   Comments: 0

calculate ∫_0 ^∞ ((x^4 dx)/((2x+1)^5 (3x+1)^8 ))

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{x}^{\mathrm{4}} \mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{8}} } \\ $$

Question Number 119752    Answers: 1   Comments: 0

find I_λ =∫_0 ^∞ ((ch(1+λcosx))/((x^2 +1)^2 ))dx (λ real >0)

$${find}\:{I}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ch}\left(\mathrm{1}+\lambda{cosx}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left(\lambda\:{real}\:>\mathrm{0}\right) \\ $$

Question Number 119750    Answers: 1   Comments: 0

Π_(k=1) ^(1019) [((2k)/(2k−1))]=?

$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{1019}} {\prod}}\left[\frac{\mathrm{2k}}{\mathrm{2k}−\mathrm{1}}\right]=? \\ $$

Question Number 119696    Answers: 2   Comments: 0

For a<b then ∫_a ^b (x−a)(x−b) dx equal to _

$${For}\:{a}<{b}\:{then}\:\underset{{a}} {\overset{{b}} {\int}}\:\left({x}−{a}\right)\left({x}−{b}\right)\:{dx}\: \\ $$$${equal}\:{to}\:\_ \\ $$

Question Number 119681    Answers: 4   Comments: 0

∫_0 ^π (√((1+cos2x)/2)) dx ∫_0 ^∞ [ne^(−x) ]dx

$$\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx} \\ $$

Question Number 119570    Answers: 3   Comments: 0

... ♣_♣ ^♣ nice calculus♣_♣ ^♣ ... prove that :: Ω=∫_0 ^( ∞) e^(−2x) ln(((1+e^(−x) )/(1−e^(−x) )))=1 ...★ M.N.1970★...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}{nice}\:\:{calculus}\underset{\clubsuit} {\overset{\clubsuit} {\clubsuit}}... \\ $$$$\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} {e}^{−\mathrm{2}{x}} {ln}\left(\frac{\mathrm{1}+{e}^{−{x}} }{\mathrm{1}−{e}^{−{x}} }\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\bigstar\:\mathscr{M}.\mathscr{N}.\mathrm{1970}\bigstar... \\ $$

Question Number 119591    Answers: 0   Comments: 6

...nice calculus... prove that :: ∫_0 ^( (π/2)) (√(((2^x −1)sin^3 (x))/((2^x +1)(sin^3 (x)+cos^3 (x))))) dx<(π/8) ...m.n.1970...

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$ \\ $$$$\:\: \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \sqrt{\frac{\left(\mathrm{2}^{{x}} −\mathrm{1}\right){sin}^{\mathrm{3}} \left({x}\right)}{\left(\mathrm{2}^{{x}} +\mathrm{1}\right)\left({sin}^{\mathrm{3}} \left({x}\right)+{cos}^{\mathrm{3}} \left({x}\right)\right)}}\:\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}.\mathrm{1970}... \\ $$$$ \\ $$

Question Number 119462    Answers: 2   Comments: 0

... advanced calculus... evaluate :: Ω=∫_0 ^( ∞) ((tan^(−1) (x))/(e^(2πx) −1))dx =? m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...\:{advanced}\:{calculus}... \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{tan}^{−\mathrm{1}} \left({x}\right)}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119446    Answers: 0   Comments: 0

calculste ∫_0 ^∞ ((ln(2+x^2 ))/(1+x^3 ))dx

$${calculste}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$

Question Number 119445    Answers: 0   Comments: 0

...nice calculus... prove that:: Σ_(n=1) ^∞ (((−1)^(n−1) )/(n^3 (((2n)),(n) ))) =^(???) ζ(3) m.n.1970

$$\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$

Question Number 119442    Answers: 0   Comments: 0

... advanced calculus... prove that : Σ_(n=1) ^∞ (1/(n^2 (((2n)),(n) ))) =^(???) ((ζ(2))/3) solution:: Σ_(n=1) ^∞ (1/(n^2 ∗(((2n)!)/((n!)^2 ))))=Σ_(n=1) ^∞ ((n!∗n!)/(n^2 ∗(2n)!)) =Σ_(n=1) ^∞ ((Γ(n)Γ(n+1))/(nΓ(2n+1)))=Σ_(n=1) ^∞ ((β(n,n+1))/n) =Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n−1) (1−x)^n =∫_0 ^( 1) (1/x)Σ(((x−x^2 )^n )/n)dx =−∫_0 ^( 1) ((ln(1−x+x^2 ))/x)dx =−∫_0 ^( 1) ((ln(1+x^3 )−ln(1+x))/x)dx =∫_0 ^( 1) ((ln(1+x))/x)dx −∫_0 ^( 1) ((ln(1+x^3 ))/x)dx =−li_2 (−1) −∫_0 ^( 1) ((Σ_(n=1) (((−1)^(n+1) x^(3n) )/n))/x) dx =(π^2 /(12))+Σ_(n=1) ^∞ (((−1)^n )/n)∫_0 ^( 1) x^(3n−1) dx =(π^2 /(12)) +(1/3)Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(12))−(π^2 /(36)) =(π^2 /(18)) =((ζ(2))/3) ✓✓ m.n.july.1970..

$$\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\::\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}\:\overset{???} {=}\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}} \\ $$$${solution}::\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \ast\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{n}!\ast{n}!}{{n}^{\mathrm{2}} \ast\left(\mathrm{2}{n}\right)!}\: \\ $$$$\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\Gamma\left({n}\right)\Gamma\left({n}+\mathrm{1}\right)}{{n}\Gamma\left(\mathrm{2}{n}+\mathrm{1}\right)}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\beta\left({n},{n}+\mathrm{1}\right)}{{n}} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{{n}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{n}} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}}\Sigma\frac{\left({x}−{x}^{\mathrm{2}} \right)^{{n}} }{{n}}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}{{x}}{dx} \\ $$$$\:\:=−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{{x}}{dx} \\ $$$$\:=−{li}_{\mathrm{2}} \left(−\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\underset{{n}=\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{3}{n}} }{{n}}}{{x}}\:{dx} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{3}{n}−\mathrm{1}} {dx}\:\:\:\:\: \\ $$$$ \\ $$$$\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\pi^{\mathrm{2}} }{\mathrm{36}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\:=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{3}}\:\:\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}.. \\ $$$$\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$

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