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.......Advanced ...∗∗∗∗∗ ...Integral...... Prove that :: Φ :=∫_0 ^( 1) ((1−x)/((1−x+x^2 )log(x)))dx= proof:: Φ:=∫_0 ^( 1) ((1−x^2 )/((1−x^3 )log(x)))dx f (a):= ∫_0 ^( 1) ((1−x^a )/((1−x^3 )log(x))) Φ := f (2) ........✓ f ′(a):=∫_0 ^( 1) ((−x^a log(x))/((1−x^3 )log(x)))=∫_0 ^( 1) ((−x^a )/(1−x^3 ))dx (★) (★):: x^3 =y ⇒ f ′(a):=(1/3)∫_0 ^( 1) ((y^((−2)/3) −y^((a/3)−(2/3)) )/(1−y))dy :=(1/3)∫_0 ^( 1) ((y^((−2)/3) −1+1−y^((a/3)−(1/3)) )/(1−y))dy :=(1/3)(ψ((a/3)+(2/3))−ψ((2/3))) f (a):=log(Γ((a/3)+(2/3)))−(a/3)ψ((2/3))+C f (0):=0=log(Γ((2/3)))+C C :=−log(Γ((2/3))) Φ:= f (2)=log(Γ((4/3)))−(2/3) ψ((2/3))−log(Γ((2/3))) :=log(((Γ((4/3)))/(Γ((2/3)))))−(2/3)ψ((2/3)) ....✓

$. . . . . . . A d v a n c e d . . . * * * * * . . . I n t e g r a l . . . . . .$ $P r o v e t h a t :: Φ := \int_{0}^{1} \frac{1 - x}{(1 - x + x^{2}) l o g (x)} d x =$ $p r o o f ::$ $Φ := \int_{0}^{1} \frac{1 - x^{2}}{(1 - x^{3}) l o g (x)} d x$ $f (a) := \int_{0}^{1} \frac{1 - x^{a}}{(1 - x^{3}) l o g (x)}$ $Φ := f (2) . . . . . . . . ✓$ $f^{'} (a) := \int_{0}^{1} \frac{- x^{a} l o g (x)}{(1 - x^{3}) l o g (x)} = \int_{0}^{1} \frac{- x^{a}}{1 - x^{3}} d x (★)$ $(★) :: x^{3} = y \Rightarrow f^{'} (a) := \frac{1}{3} \int_{0}^{1} \frac{y^{\frac{- 2}{3}} - y^{\frac{a}{3} - \frac{2}{3}}}{1 - y} d y$ $:= \frac{1}{3} \int_{0}^{1} \frac{y^{\frac{- 2}{3}} - 1 + 1 - y^{\frac{a}{3} - \frac{1}{3}}}{1 - y} d y$ $:= \frac{1}{3} (ψ (\frac{a}{3} + \frac{2}{3}) - ψ (\frac{2}{3}))$ $f (a) := l o g (Γ (\frac{a}{3} + \frac{2}{3})) - \frac{a}{3} ψ (\frac{2}{3}) + C$ $f (0) := 0 = l o g (Γ (\frac{2}{3})) + C$ $C := - l o g (Γ (\frac{2}{3}))$ $Φ := f (2) = l o g (Γ (\frac{4}{3})) - \frac{2}{3} ψ (\frac{2}{3}) - l o g (Γ (\frac{2}{3}))$ $:= l o g (\frac{Γ (\frac{4}{3})}{Γ (\frac{2}{3})}) - \frac{2}{3} ψ (\frac{2}{3}) . . . . ✓$

Question Number 142268 Answers: 0 Comments: 0

∫(e^x /(cosx))dx

$\int \frac{e^{x}}{c o s x} d x$

Question Number 141947 Answers: 2 Comments: 0

Ω:=∫_0 ^( 1) (((√(1−x)) arcsin(x))/( (√(1+x))))dx=??

$Ω := \int_{0}^{1} \frac{\sqrt{1 - x} a r c s i n (x)}{\sqrt{1 + x}} d x = ? ?$

Question Number 141916 Answers: 1 Comments: 0

prove that:: I:=∫_0 ^( (π/2)) arccosh(sin(x)+cos(x))dx=(π/2)ln(2) ..

$p r o v e t h a t ::$ $I := \int_{0}^{\frac{π}{2}} a r c c o s h (s i n (x) + c o s (x)) d x = \frac{π}{2} l n (2) . .$

Question Number 141797 Answers: 0 Comments: 0

Question Number 141759 Answers: 0 Comments: 0

Question Number 141716 Answers: 1 Comments: 0

(x^2 lnx)y′′−xy′+y=0

$(x^{2} l n x) y^{″} - {x y}^{'} + y = 0$

Question Number 141668 Answers: 1 Comments: 0

.......Advanced ...★ ...★ ... Calculus....... if Ω =Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) then prove that :: (1/2) = e^(Ω−1) proof :: method (1): ψ (1+x )= −γ+Σ_(n=2) ^∞ (−1)^n ζ(n)x^(n−1) ( Maclaurin series for ψ(x+1) ) x:=(1/2) ⇒ ψ ((3/2) )=−γ + 2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) (∗ ) we know that :: ψ(1+x)=(1/x)+ψ(x) ( ∗ ) ⇛ ψ ((3/2))=2+ψ((1/2))=−γ+2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) (∗) ⇛ 2−γ−ln(4)=−γ+2Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) ln((e/2))= Σ_(n=2) ^∞ (((−1)^n ζ(n))/2^n ) =Ω (1/2) = e^(Ω −1) ....✓ ...m.n.july.1970...

$. . . . . . . A d v a n c e d . . . ★ . . . ★ . . . C a l c u l u s . . . . . . .$ $i f Ω = \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}} t h e n p r o v e$ $t h a t :: \frac{1}{2} = e^{Ω - 1}$ $p r o o f ::$ $m e t h o d (1) :$ $ψ (1 + x) = - γ + \underset{n = 2}{\sum^{\infty}} {(- 1)}^{n} ζ (n) x^{n - 1}$ $(M a c l a u r i n s e r i e s f o r ψ (x + 1))$ $x := \frac{1}{2} \Rightarrow ψ (\frac{3}{2}) = - γ + 2 \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}} (*)$ $w e k n o w t h a t :: ψ (1 + x) = \frac{1}{x} + ψ (x)$ $(*) ⇛ ψ (\frac{3}{2}) = 2 + ψ (\frac{1}{2}) = - γ + 2 \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}}$ $(*) ⇛ 2 - γ - l n (4) = - γ + 2 \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}}$ $l n (\frac{e}{2}) = \underset{n = 2}{\sum^{\infty}} \frac{{(- 1)}^{n} ζ (n)}{2^{n}} = Ω$ $\frac{1}{2} = e^{Ω - 1} . . . . ✓$ $. . . m . n . j u l y . 1970 . . .$

Question Number 141512 Answers: 0 Comments: 0

Σ_(n=1 ) ^∞ ((sin((((2n−1)π)/6)))/((2n−1)^2 )) = a.G a = ? (G:= catalan constant)

$\underset{n = 1}{\sum^{\infty}} \frac{s i n (\frac{(2 n - 1) π}{6})}{{(2 n - 1)}^{2}} = a . G$ $a = ? (G := c a t a l a n c o n s t a n t)$

Question Number 141475 Answers: 2 Comments: 0

Find the smallest value of (√(x^2 +y^2 )) among all values of x & y satisfying 3x−y = 20

$F i n d t h e s m a l l e s t v a l u e o f$ $\sqrt{x^{2} + y^{2}} a m o n g a l l v a l u e s o f$ $x & y s a t i s f y i n g 3 x - y = 20$

Question Number 141444 Answers: 0 Comments: 0

Question Number 141346 Answers: 1 Comments: 0

Question Number 141322 Answers: 0 Comments: 0

......advanced........calculus....... prove that:: ξ:=Π_(n=2) ^∞ e(1−(1/n^2 ))^n^2 =(π/(e(√e)))

$. . . . . . a d v a n c e d . . . . . . . . c a l c u l u s . . . . . . .$ $p r o v e t h a t ::$ $ξ := \underset{n = 2}{\prod^{\infty}} e {(1 - \frac{1}{n^{2}})}^{n^{2}} = \frac{π}{e \sqrt{e}}$

Question Number 141320 Answers: 1 Comments: 0

......nice ......calculuus..... prove that:: 𝛗:=∫_0 ^( ∞) ∫_0 ^( ∞) ((A rctan(x^2 y^2 ))/(x^4 +y^4 ))dxdy=((π^2 (√2))/(16)) .....

$. . . . . . n i c e . . . . . . c a l c u l u u s . . . . .$ $p r o v e t h a t ::$ $ϕ := \int_{0}^{\infty} \int_{0}^{\infty} \frac{A r c t a n (x^{2} y^{2})}{x^{4} + y^{4}} d x d y = \frac{π^{2} \sqrt{2}}{16}$ $. . . . .$

Question Number 141368 Answers: 2 Comments: 0

A closed cylindrical can be is to hold 1 liters of liquid . How should we choose the height and radius to minimize the amount of material needed to manufacture the can ?

$A c l o s e d c y l i n d r i c a l c a n b e i s t o h o l d$ $1 l i t e r s o f l i q u i d . H o w s h o u l d w e$ $c h o o s e t h e h e i g h t a n d r a d i u s$ $t o m i n i m i z e t h e a m o u n t o f$ $m a t e r i a l n e e d e d t o m a n u f a c t u r e$ $t h e c a n ?$

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