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IntegrationQuestion and Answers: Page 110

Question Number 128529 Answers: 1 Comments: 2

Question Number 128511 Answers: 1 Comments: 0

H=∫ ((2017x^(2016) +2018x^(2017) )/(1+x^(4034) +2x^(4035) +x^(4036) )) dx

$H = \int \frac{{2017 x}^{2016} + {2018 x}^{2017}}{1 + x^{4034} + {2 x}^{4035} + x^{4036}} dx$

Question Number 128499 Answers: 1 Comments: 0

find u_n =∫_1 ^∞ (([ne^(−x) ])/n^3 )dx

$find u_{n} = \int_{1}^{\infty} \frac{[{ne}^{- x}]}{n^{3}} dx$

Question Number 128495 Answers: 0 Comments: 0

find ∫_0 ^∞ ((sinx)/([x]))dx

$find \int_{0}^{\infty} \frac{sinx}{[x]} dx$

Question Number 128471 Answers: 2 Comments: 0

... calculus ... Φ=^? ∫_0 ^( 1) (ln(x))^2 ln((√(−ln(x))) dx

$. . . c a l c u l u s . . .$ $Φ \overset{?}{=} \int_{0}^{1} {(l n (x))}^{2} l n (\sqrt{- l n (x)} d x$

Question Number 128417 Answers: 1 Comments: 0

η = ∫_0 ^( 1) x^3 (1−x^3 )^(n−1) dx

$η = \int_{0}^{1} x^{3} {(1 - x^{3})}^{n - 1} dx$

Question Number 128408 Answers: 1 Comments: 0

ρ = ∫ ((sin (4x))/(sin^4 (x)+cos^4 (x))) dx

$ρ = \int \frac{\sin (4 x)}{\sin^{4} (x) + \cos^{4} (x)} d x$

Question Number 128385 Answers: 1 Comments: 0

∫_1 ^( π) determinant ((x^3 ,(lnx),(sinx)),((3x^2 ),(1/x),(cosx)),(6,(2x^(−3) ),(−cosx)))dx =?

$\int_{1}^{π} | \begin{matrix} x^{3} & lnx & sinx \\ {3 x}^{2} & \frac{1}{x} & cosx \\ 6 & {2 x}^{- 3} & - cosx \end{matrix} | dx = ?$

Question Number 128373 Answers: 1 Comments: 0

∫_0 ^1 x^(3/2) (1−x)^(1/2) dx

$\int_{0}^{1} x^{\frac{3}{2}} {(1 - x)}^{\frac{1}{2}} d x$

Question Number 128334 Answers: 1 Comments: 0

Ω = ∫ (x^2 /( (√((a+bx^2 )^5 )))) dx ; where : a; b >0

$Ω = \int \frac{x^{2}}{\sqrt{{(a + {bx}^{2})}^{5}}} dx; where : a; b > 0$

Question Number 128346 Answers: 2 Comments: 0

Question Number 128316 Answers: 1 Comments: 0

nice calculus Ω= ∫_0 ^( ∞) ((sin^3 (x))/x^2 )dx=?

$n i c e c a l c u l u s$ $Ω = \int_{0}^{\infty} \frac{{s i n}^{3} (x)}{x^{2}} d x = ?$

Question Number 128285 Answers: 1 Comments: 0

cos ((π/7))−cos (((2π)/7))+cos (((3π)/7)) =?

$\cos (\frac{π}{7}) - \cos (\frac{2 π}{7}) + \cos (\frac{3 π}{7}) = ?$

Question Number 128276 Answers: 1 Comments: 0

∫_e^2 ^( ∞) (dx/(x^3 ln x)) ?

$\int_{e^{2}}^{\infty} \frac{d x}{x^{3} \ln x} ?$

Question Number 128262 Answers: 2 Comments: 0

Question Number 128251 Answers: 1 Comments: 0

Question Number 128244 Answers: 0 Comments: 0

...nice calculus... prove that ::Ω= ∫_0 ^(π/4) ln(sin(x))d=((−π)/4)log(2)−(G/2) log(2sin(x))=Σ_(n=1) ^∞ ((−1)/n)cos(2nx) Ω= ∫_0 ^( (π/4)) {−log(2)−Σ_(n=1) ^∞ ((cos(2nx))/n)}dx =((−π)/4)log(2)−Σ_(n=1) ^∞ ∫_0 ^( (π/4)) ((cos(2nx))/n)dx =((−π)/4)log(2)−Σ_(n=1) ^∞ [(1/(2n^2 ))sin(2nx)]_0 ^(π/4) =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ ((sin(((nπ)/2)))/n^2 ) =((−π)/4)log(2)−(1/2){(1/1^2 )−(1/3^2 ) +(1/5^2 )−..} =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 )) ∴ Ω =((−π)/4)log(2)−(G/2) ✓ G:= catalan constant...

$. . . n i c e c a l c u l u s . . .$ $p r o v e t h a t :: Ω = \int_{0}^{\frac{π}{4}} l n (s i n (x)) d = \frac{- π}{4} l o g (2) - \frac{G}{2}$ $l o g (2 s i n (x)) = \underset{n = 1}{\sum^{\infty}} \frac{- 1}{n} c o s (2 n x)$ $Ω = \int_{0}^{\frac{π}{4}} {- l o g (2) - \underset{n = 1}{\sum^{\infty}} \frac{c o s (2 n x)}{n}} d x$ $= \frac{- π}{4} l o g (2) - \underset{n = 1}{\sum^{\infty}} \int_{0}^{\frac{π}{4}} \frac{c o s (2 n x)}{n} d x$ $= \frac{- π}{4} l o g (2) - \underset{n = 1}{\sum^{\infty}} {[\frac{1}{2 n^{2}} s i n (2 n x)]}_{0}^{\frac{π}{4}}$ $= \frac{- π}{4} l o g (2) - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{s i n (\frac{n π}{2})}{n^{2}}$ $= \frac{- π}{4} l o g (2) - \frac{1}{2} {\frac{1}{1^{2}} - \frac{1}{3^{2}} + \frac{1}{5^{2}} - . .}$ $= \frac{- π}{4} l o g (2) - \frac{1}{2} \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{{(2 n - 1)}^{2}}$ $∴ Ω = \frac{- π}{4} l o g (2) - \frac{G}{2} ✓$ $G := c a t a l a n c o n s t a n t . . .$