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IntegrationQuestion and Answers: Page 95

Question Number 135697 Answers: 0 Comments: 0

Question Number 135693 Answers: 1 Comments: 0

Ω = ∫ ((x−1)/((x−2)(x^2 −2x+2)^2 )) dx

$Ω = \int \frac{x - 1}{(x - 2) {(x^{2} - 2 x + 2)}^{2}} d x$

Question Number 135673 Answers: 2 Comments: 0

∫ ((x^2 +1)/(x^4 +x^2 +1)) dx

$\int \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x$

Question Number 135646 Answers: 1 Comments: 0

∫(x+(1/2))ln(1+(1/x))−x dx=...?

$\int (x + \frac{1}{2}) l n (1 + \frac{1}{x}) - x d x = . . . ?$

Question Number 135633 Answers: 1 Comments: 0

∫(1/(x^(1/3) +1))dx=...?

$\int \frac{1}{x^{\frac{1}{3}} + 1} d x = . . . ?$

Question Number 135662 Answers: 0 Comments: 0

Question Number 135627 Answers: 1 Comments: 0

... nice ................. calculus ... evaluation::::: 𝛗=^(???) ∫_0 ^( (π/2)) sin(x)ln(sin(x))dx solution::::: 𝛗=^(⟨cos(x)=y⟩) (1/2)∫_0 ^( 1) ln(1−y^2 )dy =−(1/2)∫_0 ^( 1) Σ_(n=1 ) ^∞ (y^(2n) /n)=((−1)/2)Σ_(n=1) ^∞ ((1/n)∫_0 ^( 1) y^(2n) dy) =((−1)/2)Σ_(n=1) ^∞ (1/(n(2n+1)))=−Σ_(n=1) ^∞ (1/(2n)) −(1/(2n+1)) =−((1/2)−(1/3)+(1/4)−(1/5)+...) =−1+(1−(1/2)+(1/3)−...) =−1+Σ_(n=1) ^∞ (((−1)^(n−1) )/n)=_(harmonic seties) ^(alternating) −1+ln(2) ∴ 𝛗= −1+ln(2)=ln((2/e))

$. . . n i c e . . . . . . . . . . . . . . . . . c a l c u l u s . . .$ $e v a l u a t i o n ::::: ϕ \overset{? ? ?}{=} \int_{0}^{\frac{π}{2}} s i n (x) l n (s i n (x)) d x$ $s o l u t i o n :::::$ $ϕ \overset{⟨ c o s (x) = y ⟩}{=} \frac{1}{2} \int_{0}^{1} l n (1 - y^{2}) d y$ $= - \frac{1}{2} \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{y^{2 n}}{n} = \frac{- 1}{2} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} \int_{0}^{1} y^{2 n} d y)$ $= \frac{- 1}{2} \underset{n = 1}{\sum^{\infty}} \frac{1}{n (2 n + 1)} = - \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n} - \frac{1}{2 n + 1}$ $= - (\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + . . .)$ $= - 1 + (1 - \frac{1}{2} + \frac{1}{3} - . . .)$ $= - 1 + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} \underset{h a r m o n i c s e t i e s}{\overset{a l t e r n a t i n g}{=}} - 1 + l n (2)$ $∴ ϕ = - 1 + l n (2) = l n (\frac{2}{e})$

Question Number 135614 Answers: 0 Comments: 0

Question Number 135610 Answers: 1 Comments: 0

.... Advanced ...... Calculus.... prove that : determinant (((i :: Π_(n=0) ^∞ (1−(x^2 /((2n+1)^2 ))) =cos(((πx)/2)) ✓ )),((ii :: Π_(n=0) ^∞ (1+(x^2 /((2n+1)^2 )))= cosh(((πx)/2)) ✓✓))) .............

$. . . . A d v a n c e d . . . . . . C a l c u l u s . . . .$ $p r o v e t h a t :$ $\begin{array}{cc} i :: \underset{n = 0}{\prod^{\infty}} (1 - \frac{x^{2}}{{(2 n + 1)}^{2}}) = c o s (\frac{π x}{2}) ✓ \\ i i :: \underset{n = 0}{\prod^{\infty}} (1 + \frac{x^{2}}{{(2 n + 1)}^{2}}) = c o s h (\frac{π x}{2}) ✓ ✓ \end{array}$ $. . . . . . . . . . . . .$

Question Number 135559 Answers: 2 Comments: 1

Question Number 135525 Answers: 0 Comments: 2

.... nice ................ calculus... evaluation of :: 𝛗=∫_0 ^( ∞) xe^(−x) (√(1−e^(−x) )) dx solution:: 1−e^(−x) =t ⇒ {_( x=−ln(1−t)) ^( e^(−x) dx=dt) 𝛗=−∫_0 ^( 1) ln(1−t).t^(1/2) dt =∫_0 ^( 1) Σ_(n=1) ^∞ (t^(n+(1/2)) /n)dt=Σ_(n=1) ^∞ (1/(n(n+(3/2)))) .... ∴ 𝛗= Σ_(n=1) ^∞ (1/(n(n+(3/2))))=Σ_(n=1) ^∞ (1/n)−(1/(n+(3/2))) .... =(2/3){γ−γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+(3/2))))} .... we know that : ψ(s+1) := −γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+s))) ..... ∴ 𝛗=(2/3)(γ+ψ((5/2))) .... on the oyher hand we have:: ψ(s+1)=(1/s)+ψ(s) ...... ψ((1/2))=−γ−2ln(2) ...... ψ((5/2))=(2/3)+ψ((3/2))=(2/3)+(2+ψ((1/2))) =(2/3)+2+(−γ−2ln(2)) .... ∴ ψ((3/2))=(8/3)−γ−ln(4) .... :::::: 𝛗 =(2/3)(γ+(8/3)−γ−ln(4)) .... 𝛗=(2/3)((8/3)−ln(4))=(4/3)((4/3)−ln(2)) .... ...... 𝛗= (4/3)((4/3)−ln(2))......✓✓ .... ... prepare by mr rizzy−aka... solution with detais :m.n.july.1970

$. . . . n i c e . . . . . . . . . . . . . . . . c a l c u l u s . . .$ $e v a l u a t i o n o f :: ϕ = \int_{0}^{\infty} {x e}^{- x} \sqrt{1 - e^{- x}} d x$ $s o l u t i o n ::$ $1 - e^{- x} = t \Rightarrow {_{x = - l n (1 - t)}^{e^{- x} d x = d t}$ $ϕ = - \int_{0}^{1} l n (1 - t) . t^{\frac{1}{2}} d t$ $= \int_{0}^{1} \underset{n = 1}{\sum^{\infty}} \frac{t^{n + \frac{1}{2}}}{n} d t = \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + \frac{3}{2})} . . . .$ $∴ ϕ = \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + \frac{3}{2})} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \frac{1}{n + \frac{3}{2}} . . . .$ $= \frac{2}{3} {γ - γ + \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + \frac{3}{2}})} . . . .$ $w e k n o w t h a t :$ $ψ (s + 1) := - γ + \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + s}) . . . . .$ $∴ ϕ = \frac{2}{3} (γ + ψ (\frac{5}{2})) . . . .$ $o n t h e o y h e r h a n d w e h a v e ::$ $ψ (s + 1) = \frac{1}{s} + ψ (s) . . . . . .$ $ψ (\frac{1}{2}) = - γ - 2 l n (2) . . . . . .$ $ψ (\frac{5}{2}) = \frac{2}{3} + ψ (\frac{3}{2}) = \frac{2}{3} + (2 + ψ (\frac{1}{2}))$ $= \frac{2}{3} + 2 + (- γ - 2 l n (2)) . . . .$ $∴ ψ (\frac{3}{2}) = \frac{8}{3} - γ - l n (4) . . . .$ $:::::: ϕ = \frac{2}{3} (γ + \frac{8}{3} - γ - l n (4)) . . . .$ $ϕ = \frac{2}{3} (\frac{8}{3} - l n (4)) = \frac{4}{3} (\frac{4}{3} - l n (2)) . . . .$ $. . . . . . ϕ = \frac{4}{3} (\frac{4}{3} - l n (2)) . . . . . . ✓ ✓ . . . .$ $. . . p r e p a r e b y m r r i z z y - a k a . . .$ $s o l u t i o n w i t h d e t a i s : m . n . j u l y . 1970$

Question Number 135513 Answers: 1 Comments: 0

Question Number 135495 Answers: 1 Comments: 0

hi, guyz ! let′s try this : I=∫_0 ^( 1) ((sin^2 x)/(cos^3 x))dx.

$hi, guyz!$ ${let}^{'} s try this : I = \int_{0}^{1} \frac{{s i n}^{2} x}{{c o s}^{3} x} d x .$

Question Number 135443 Answers: 0 Comments: 0

Question Number 135389 Answers: 0 Comments: 0

let U_n =∫_(−∞) ^∞ ((cos(nx))/((x^2 −x+1)^2 ))dx calculate lim_(n→∞) e^n^2 U_n

$l e t U_{n} = \int_{- \infty}^{\infty} \frac{c o s (n x)}{{(x^{2} - x + 1)}^{2}} d x$ $c a l c u l a t e {l i m}_{n \to \infty} e^{n^{2}} U_{n}$

Question Number 135382 Answers: 1 Comments: 0

find ∫_0 ^1 x^n ln(1−x^4 )dx with n integr natural

$f i n d \int_{0}^{1} x^{n} l n (1 - x^{4}) d x w i t h n$ $i n t e g r n a t u r a l$

Question Number 135372 Answers: 0 Comments: 0

calculate ∫_0 ^∞ (dx/(((√x)+(√(1+x^2 )))^3 ))

$c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(\sqrt{x} + \sqrt{1 + x^{2}})}^{3}}$

Question Number 135368 Answers: 1 Comments: 0

calculate ∫_0 ^(+∞) ((xarctan(2x))/((x^2 +1)^2 ))dx

$c a l c u l a t e \int_{0}^{+ \infty} \frac{x a r c t a n (2 x)}{{(x^{2} + 1)}^{2}} d x$

Question Number 135361 Answers: 1 Comments: 0

f(x)=3x^2 +6x,[−1,5] find−the−average−value

$f (x) = 3 x^{2} + 6 x, [- 1, 5]$ $f i n d - t h e - a v e r a g e - v a l u e$

Question Number 135259 Answers: 2 Comments: 0

∫(x^2 +3x)cos (x)dx

$\int (x^{2} + 3 x) \cos (x) d x$

Question Number 135257 Answers: 2 Comments: 1

∫t^7 ln(t)dt

$\int t^{7} l n (t) d t$

Question Number 135215 Answers: 0 Comments: 0

.....calculus preliminary.... Q: f(x)=2^x −2^(−x) ⇒ f^( −1) (x)=??? solution: y=2^x −2^(−x) ..... y=((2^(2x) −1)/2^x ) ⇒2^(2x) −y2^x −1=0 (∗) ... :: 2^x =t⇒ t>0 ...✓ .... (∗)→... t^2 −ty−1=0 .... Δ=y^2 +4>0...✓ ... t=((y+(√(y^2 +4)))/2) ...... :: 2^x =((y+(√(y^2 +4)))/2) ⇒_(both sides) ^(taking log) .... x:=log_2 (((y+(√(y^2 +4)))/2)) f^( −1) (x)=log_2 (((x+(√(x^2 +4)))/2)) ✓✓ .........................

$. . . . . c a l c u l u s p r e l i m i n a r y . . . .$ $Q : f (x) = 2^{x} - 2^{- x} \Rightarrow f^{- 1} (x) = ? ? ?$ $s o l u t i o n :$ $y = 2^{x} - 2^{- x} . . . . .$ $y = \frac{2^{2 x} - 1}{2^{x}} \Rightarrow 2^{2 x} - y 2^{x} - 1 = 0 (*) . . .$ $:: 2^{x} = t \Rightarrow t > 0 . . . ✓ . . . .$ $(*) \to . . . t^{2} - t y - 1 = 0 . . . .$ $Δ = y^{2} + 4 > 0 . . . ✓ . . .$ $t = \frac{y + \sqrt{y^{2} + 4}}{2} . . . . . .$ $:: 2^{x} = \frac{y + \sqrt{y^{2} + 4}}{2} \underset{b o t h s i d e s}{\overset{t a k i n g l o g}{\Rightarrow}} . . . .$ $x := {l o g}_{2} (\frac{y + \sqrt{y^{2} + 4}}{2})$ $f^{- 1} (x) = {l o g}_{2} (\frac{x + \sqrt{x^{2} + 4}}{2}) ✓ ✓$ $. . . . . . . . . . . . . . . . . . . . . . . . .$

Question Number 135174 Answers: 1 Comments: 0

Z = ∫_0 ^( π/2) arctan (sin x) dx + ∫_0 ^( π/4) arcsin (tan x) dx

$Z = \int_{0}^{π / 2} \arctan (\sin x) dx + \int_{0}^{π / 4} \arcsin (\tan x) dx$

Question Number 135231 Answers: 0 Comments: 0

∫(√((x^2 +x)^3 )) dx help me

$\int \sqrt{{(x^{2} + x)}^{3}} dx$ $help me$

Question Number 135103 Answers: 0 Comments: 1

f(x)=1+Σ_(n=2) ^∞ (((−x)^n )/n)

$f (x) = 1 + \underset{n = 2}{\sum^{\infty}} \frac{{(- x)}^{n}}{n}$

Question Number 135062 Answers: 1 Comments: 0

Let f(0) = a ; f(3)=0 and f ′(x)=e^x^4 what is the value ∫_0 ^( 3) x^2 f(x) dx ?

$Let f (0) = a; f (3) = 0 and f^{'} (x) = e^{x^{4}}$ $what is the value \int_{0}^{3} x^{2} f (x) dx ?$

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