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IntegrationQuestion and Answers: Page 126

Question Number 119681 Answers: 4 Comments: 0

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

$\int_{0}^{π} \sqrt{\frac{1 + c o s 2 x}{2}} d x$ $\int_{0}^{\infty} [{n e}^{- x}] d x$

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★...

$. . . \underset{♣}{\overset{♣}{♣}} n i c e c a l c u l u s \underset{♣}{\overset{♣}{♣}} . . .$ $p r o v e t h a t ::$ $Ω = \int_{0}^{\infty} e^{- 2 x} l n (\frac{1 + e^{- x}}{1 - e^{- x}}) = 1$ $. . . ★ M . N . 1970 ★ . . .$

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...

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $\int_{0}^{\frac{π}{2}} \sqrt{\frac{(2^{x} - 1) {s i n}^{3} (x)}{(2^{x} + 1) ({s i n}^{3} (x) + {c o s}^{3} (x))}} d x < \frac{π}{8}$ $. . . m . n . 1970 . . .$

Question Number 119462 Answers: 2 Comments: 0

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

$. . . a d v a n c e d c a l c u l u s . . .$ $e v a l u a t e ::$ $Ω = \int_{0}^{\infty} \frac{{t a n}^{- 1} (x)}{e^{2 π x} - 1} d x = ?$ $m . n . 1970$

Question Number 119446 Answers: 0 Comments: 0

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

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

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

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n^{3} (\begin{matrix} 2 n \\ n \end{matrix})} \overset{? ? ?}{=} ζ (3)$ $m . n . 1970$

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..

Question Number 119425 Answers: 1 Comments: 0

decompose F(x) =((2x−1)/((x^2 −1)^2 (x^2 +3))) and calculate ∫_(√2) ^(+∞) F(x)dx

$decompose F (x) = \frac{2 x - 1}{{(x^{2} - 1)}^{2} (x^{2} + 3)}$ $and calculate \int_{\sqrt{2}}^{+ \infty} F (x) dx$

Question Number 119335 Answers: 2 Comments: 0

Express f(x) = (1/((x−1)^2 (x^2 +1))) into partial fractions. hence evaluate I = ∫_0 ^4 f(x) dx

$Express f (x) = \frac{1}{{(x - 1)}^{2} (x^{2} + 1)} into partial fractions .$ $hence evaluate I = \underset{0}{\int^{4}} f (x) d x$

Question Number 119323 Answers: 1 Comments: 4

Question Number 119306 Answers: 3 Comments: 0

Question Number 119282 Answers: 1 Comments: 0

... nice calculus... evaluate:: I:= ∫_0 ^( 1) li_2 (1−x^2 )dx=?? .m.n.1970.

$. . . n i c e c a l c u l u s . . .$ $e v a l u a t e ::$ $I := \int_{0}^{1} {l i}_{2} (1 - x^{2}) d x = ? ?$ $. m . n . 1970 .$

Question Number 119318 Answers: 0 Comments: 0

find ∫_0 ^∞ ((lnx)/(x^4 +x^2 +2))dx

$f i n d \int_{0}^{\infty} \frac{l n x}{x^{4} + x^{2} + 2} d x$

Question Number 119317 Answers: 0 Comments: 0

calculate ∫_0 ^(2π) (dθ/((x^2 −2cosθ x+1)^2 ))

$c a l c u l a t e \int_{0}^{2 π} \frac{d θ}{{(x^{2} - 2 c o s θ x + 1)}^{2}}$

Question Number 119262 Answers: 3 Comments: 0

∫ (x^2 /( (√((4−x^2 )^5 )))) dx

$\int \frac{x^{2}}{\sqrt{{(4 - x^{2})}^{5}}} d x$

Question Number 119256 Answers: 0 Comments: 0

Determine ∫_(−(π/4)) ^( (π/4)) (cost+(√(1+t^2 sin^3 tcos^3 t))dt=?

$D e t e r m i n e \int_{- \frac{π}{4}}^{\frac{π}{4}} (c o s t + \sqrt{1 + t^{2} {s i n}^{3} {t c o s}^{3} t} d t = ?$

Question Number 119151 Answers: 0 Comments: 0

Question Number 119070 Answers: 2 Comments: 0

∫ ((x^2 −x+6)/(x^3 +3x)) dx ∫ ((5x^2 +3x−2)/(x^3 +2x^2 )) dx

$\int \frac{x^{2} - x + 6}{x^{3} + 3 x} d x$ $\int \frac{5 x^{2} + 3 x - 2}{x^{3} + 2 x^{2}} d x$

Question Number 119038 Answers: 2 Comments: 0

∫_(−π/4) ^(+π/4) ((√(1+tan x))/( (√(1−tan x))))dx

$\underset{- π / 4}{\int^{+ π / 4}} \frac{\sqrt{1 + \tan x}}{\sqrt{1 - \tan x}} d x$

Question Number 119021 Answers: 1 Comments: 0

Question Number 119006 Answers: 2 Comments: 0

Question Number 118997 Answers: 3 Comments: 0

Question Number 118959 Answers: 2 Comments: 0

∫ (dx/(x^6 −x^3 )) ?

$\int \frac{d x}{x^{6} - x^{3}} ?$

Question Number 118955 Answers: 1 Comments: 0

find ∫_0 ^∞ ((lnx)/(x^2 +x+1))dx

$f i n d \int_{0}^{\infty} \frac{l n x}{x^{2} + x + 1} d x$

Question Number 118949 Answers: 0 Comments: 0

find ∫_0 ^∞ ((sin(3cosx))/((x^2 +4)^2 ))dx

$f i n d \int_{0}^{\infty} \frac{s i n (3 c o s x)}{{(x^{2} + 4)}^{2}} d x$

Question Number 118934 Answers: 0 Comments: 0

I=∫_0 ^∞ ((lnx)/(x^2 +2x+4)) dx put x+1=(√3) tanθ I=∫_0 ^(π/2) ((ln((√3) tanθ−1))/(3(sec^2 θ))) (√3) sec^2 θdθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)tanθ−1) dθ I=(1/( (√3)))∫_0 ^(π/2) ln((√3)sinθ−cosθ)−lncosθ dθ I=(1/( (√3)))∫_0 ^(π/2) ln(sin(θ−(π/6))+ln2−lncosθ dθ A=∫_(−2) ^6 ∣g(x)∣ dx ⇒let g(p)=0⇒ f(0)=p=2 A=∫_(−2) ^2 −g(x) dx+∫_2 ^6 g(x) dx ⇒put x=f(t) A=∫_(−1) ^0 −tf′(t) dt+∫_0 ^1 t f(t) dt A=∫_(−1) ^0 −3t^3 −3t dt+∫_0 ^1 3t^3 +3t dt A=2(((3t^4 )/4)+((3t^2 )/2))_0 ^1 =(9/2) lnx=(1/x)⇒xlnx=1(let x=α) ∫_1 ^α (1/x)−lnx dx=∫_α ^a lnx−(1/x) dx ⇒lnx−xlnx+x]_1 ^α =xlnx−x−lnx]_α ^a ⇒lnα−1+α−1=alna−a−lna−1+α +lnα ⇒alna−a−lna=−1 ⇒alna−lna=a−1⇒a=e s(t)=(1/2)∣OA^→ ×OB^→ ∣ s(t)=(1/2)∣(2,2,1)×(t,1,t+1)∣ s(t)=(1/2)∣(2t+1),(−2−t),(2−2t)∣ f(x)=(1/4)∫_0 ^x (2t+1)^2 +(t+2)^2 +(2−2t)^2 dt f(x)=(1/4)∫_0 ^x 9t^2 +9 dt=((3x^3 +9x)/4) A=(1/4)∫_0 ^6 (3x^3 +9x)dx=((3x^4 +18x^2 )/(16))]_0 ^6 A=((3.6^3 (6+1))/(16))=((567)/2)

$I = \int_{0}^{\infty} \frac{l n x}{x^{2} + 2 x + 4} d x$ $p u t x + 1 = \sqrt{3} t a n θ$ $I = \int_{0}^{π / 2} \frac{l n (\sqrt{3} t a n θ - 1)}{3 ({s e c}^{2} θ)} \sqrt{3} {s e c}^{2} θ d θ$ $I = \frac{1}{\sqrt{3}} \int_{0}^{π / 2} l n (\sqrt{3} t a n θ - 1) d θ$ $I = \frac{1}{\sqrt{3}} \int_{0}^{π / 2} l n (\sqrt{3} s i n θ - c o s θ) - l n c o s θ d θ$ $I = \frac{1}{\sqrt{3}} \int_{0}^{π / 2} l n (s i n (θ - \frac{π}{6}) + l n 2 - l n c o s θ d θ$ $A = \int_{- 2}^{6} ∣ g (x) ∣ d x$ $\Rightarrow l e t g (p) = 0 \Rightarrow f (0) = p = 2$ $A = \int_{- 2}^{2} - g (x) d x + \int_{2}^{6} g (x) d x$ $\Rightarrow p u t x = f (t)$ $A = \int_{- 1}^{0} - {t f}^{'} (t) d t + \int_{0}^{1} t f (t) d t$ $A = \int_{- 1}^{0} - 3 t^{3} - 3 t d t + \int_{0}^{1} 3 t^{3} + 3 t d t$ $A = 2 {(\frac{3 t^{4}}{4} + \frac{3 t^{2}}{2})}_{0}^{1} = \frac{9}{2}$ $l n x = \frac{1}{x} \Rightarrow x l n x = 1 (l e t x = α)$ $\int_{1}^{α} \frac{1}{x} - l n x d x = \int_{α}^{a} l n x - \frac{1}{x} d x$ ${\Rightarrow {l n x - x l n x + x]}_{1}^{α} = x l n x - x - l n x]}_{α}^{a}$ $\Rightarrow l n α - 1 + α - 1 = a l n a - a - l n a - 1 + α$ $+ l n α$ $\Rightarrow a l n a - a - l n a = - 1$ $\Rightarrow a l n a - l n a = a - 1 \Rightarrow a = e$ $s (t) = \frac{1}{2} ∣ O \vec{A} \times O \vec{B} ∣$ $s (t) = \frac{1}{2} ∣ (2, 2, 1) \times (t, 1, t + 1) ∣$ $s (t) = \frac{1}{2} ∣ (2 t + 1), (- 2 - t), (2 - 2 t) ∣$ $f (x) = \frac{1}{4} \int_{0}^{x} {(2 t + 1)}^{2} + {(t + 2)}^{2} + {(2 - 2 t)}^{2} d t$ $f (x) = \frac{1}{4} \int_{0}^{x} 9 t^{2} + 9 d t = \frac{3 x^{3} + 9 x}{4}$ ${A = \frac{1}{4} \int_{0}^{6} (3 x^{3} + 9 x) d x = \frac{3 x^{4} + 18 x^{2}}{16}]}_{0}^{6}$ $A = \frac{3 . 6^{3} (6 + 1)}{16} = \frac{567}{2}$

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