Question and Answers Forum

IntegrationQuestion and Answers: Page 63

Question Number 153875 Answers: 0 Comments: 0

Prove that.. 𝛗 : =∫_( 1) ^( +∞) (( ln (x ))/(( x^( π) −1 )( ln^( 2) (x) +1 )^2 ))dx= ((π^( 2) − 8)/(16)) ■

$Prove that . .$ $ϕ : = \int_{1}^{+ \infty} \frac{l n (x)}{(x^{π} - 1) {({l n}^{2} (x) + 1)}^{2}} d x = \frac{π^{2} - 8}{16} ◼$

Question Number 153873 Answers: 0 Comments: 1

∫_0 ^( ∞) (( x)/((1 +x^( 2) ) ( e^( 2πx) − 1))) dx =((2γ− 1)/4)

$\int_{0}^{\infty} \frac{x}{(1 + x^{2}) (e^{2 π x} - 1)} d x = \frac{2 γ - 1}{4}$

Question Number 153759 Answers: 0 Comments: 0

Ω= Σ_(n=1) ^∞ {n^2 (∫_0 ^( (π/2)) (( sin^( 2) (x ))/((sin(x)+cos(x))^( 4) )))^( n) dx}=?

$Ω = \underset{n = 1}{\sum^{\infty}} {n^{2} {(\int_{0}^{\frac{π}{2}} \frac{{s i n}^{2} (x)}{{(s i n (x) + c o s (x))}^{4}})}^{n} d x} = ?$

Question Number 153734 Answers: 0 Comments: 0

Question Number 153721 Answers: 2 Comments: 0

prove that : I:= ∫_0 ^( ∞) (( x^( 3) )/(sinh ( x ))) dx = ((π^4 )/8) ■ m.n

$prove that :$ $I := \int_{0}^{\infty} \frac{x^{3}}{s i n h (x)} d x = \frac{π^{4}}{8} ◼ m . n$

Question Number 153555 Answers: 1 Comments: 0

Ω := ∫_0 ^( (π/2)) cos(2x).ln(sin(x))dx=^? −(π/4) solution (1 ) Ω := ∫_0 ^( (π/2)) ( 2cos^( 2) (x)−1)ln(sin(x))dx := 2∫_0 ^( (π/2)) cos^( 2) (x).ln(sin(x))dx−∫_0 ^( (π/2)) ln(sin(x))dx we know that : ∫_0 ^(π/2) ln(sin(x))dx=_(earlier) ^(derived) ((−π)/2) ln(2) ∫_0 ^( (π/2)) cos^( 2) (x).ln(sin(x))dx=_(posts) ^(previous) −(π/4)ln(2)−(π/8) ∴ Ω := −(π/2) ln(2) −(π/4) +(π/2) ln(2) ◂ Ω =− (π/4) ▶ m.n

$Ω := \int_{0}^{\frac{π}{2}} c o s (2 x) . l n (s i n (x)) d x \overset{?}{=} - \frac{π}{4}$ $s o l u t i o n (1)$ $Ω := \int_{0}^{\frac{π}{2}} (2 {c o s}^{2} (x) - 1) l n (s i n (x)) d x$ $:= 2 \int_{0}^{\frac{π}{2}} {c o s}^{2} (x) . l n (s i n (x)) d x - \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x$ $w e k n o w t h a t : \int_{0}^{\frac{π}{2}} l n (s i n (x)) d x \underset{e a r l i e r}{\overset{d e r i v e d}{=}} \frac{- π}{2} l n (2)$ $\int_{0}^{\frac{π}{2}} {c o s}^{2} (x) . l n (s i n (x)) d x \underset{p o s t s}{\overset{p r e v i o u s}{=}} - \frac{π}{4} l n (2) - \frac{π}{8}$ $∴ Ω := - \frac{π}{2} l n (2) - \frac{π}{4} + \frac{π}{2} l n (2)$ $◂ Ω = - \frac{π}{4} ▸ m . n$

Question Number 153535 Answers: 1 Comments: 0

find ∫((√(x^2 −9))/x^3 ) dx=?

$f i n d \int \frac{\sqrt{x^{2} - 9}}{x^{3}} d x = ?$

Question Number 153430 Answers: 1 Comments: 2

Question Number 153200 Answers: 0 Comments: 0

∫_0 ^(e−1) ∫_0 ^(e−x−1) ∫_0 ^(x+y+e) ((ln(z−x−y))/((x−e)(x+y−e)))dxdydz=?

$\int_{0}^{e - 1} \int_{0}^{e - x - 1} \int_{0}^{x + y + e} \frac{l n (z - x - y)}{(x - e) (x + y - e)} d x d y d z = ?$

Question Number 153151 Answers: 2 Comments: 0

∫_0 ^1 ((ln(x^3 +1))/(x+1))dx

$\int_{0}^{1} \frac{l n (x^{3} + 1)}{x + 1} d x$

Question Number 153117 Answers: 1 Comments: 0

∫_0 ^1 cot^(−1) (1−x+x^2 )dx

$\int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) dx$

Question Number 153114 Answers: 2 Comments: 0

∫_0 ^1 ((log (1+x))/(1+x^2 ))dx

$\int_{0}^{1} \frac{\log (1 + x)}{1 + x^{2}} dx$

Question Number 153112 Answers: 1 Comments: 0

∫_(−π) ^π (sin^(75) x+x^(125) )dx=0

$\int_{- π}^{π} (\sin^{75} x + x^{125}) dx = 0$

Question Number 153111 Answers: 0 Comments: 0

∫_(−1) ^1 e^x dx as limit of the sum

$\int_{- 1}^{1} e^{x} dx as limit of the sum$

Question Number 153110 Answers: 0 Comments: 0

∫_0 ^1 (3x^2 +2x+1)dx as the limit of sum

$\int_{0}^{1} ({3 x}^{2} + 2 x + 1) dx$ $as the limit of sum$

Question Number 153104 Answers: 1 Comments: 0

∫_0 ^π (x^3 /(x^3 +(π−x)^3 ))dx=?

$\int_{0}^{π} \frac{x^{3}}{x^{3} + {(π - x)}^{3}} d x = ?$

Question Number 153063 Answers: 0 Comments: 5

show that ∫_(−∞) ^( ∞) (1/( (√(x^2 +1)))) dx is unsolvable

$show that$ $\int_{- \infty}^{\infty} \frac{1}{\sqrt{x^{2} + 1}} d x$ $is unsolvable$

Question Number 153040 Answers: 0 Comments: 0

prove that.. Ω =∫_0 ^( ∞) (( sin (x ))/(sinh(x)))dx =(π/2) tanh ((π/2))

$p r o v e t h a t . .$ $Ω = \int_{0}^{\infty} \frac{s i n (x)}{s i n h (x)} d x = \frac{π}{2} t a n h (\frac{π}{2})$

Question Number 152939 Answers: 1 Comments: 0

Q : If a , b are positive numbers and { (( a = 1 + (( 6a −2))^(1/3) )),(( b = 1 + (( 6b −2))^(1/3) )) :} then find the value of , a.b =? ... Compiled by m.n : (E lementary olympiad ). ■

$Q : If a, b are positive numbers and$ ${\begin{cases} a = 1 + \sqrt[3]{6 a - 2} \\ b = 1 + \sqrt[3]{6 b - 2} \end{cases}$ $then find the value of, a . b = ?$ $. . . Compiled by m . n : (E l e m e n t a r y o l y m p i a d) . ◼$

Question Number 152910 Answers: 0 Comments: 0

∫^ x^x^x dx=

$\int^{} x^{x^{x}} d x =$

Question Number 152874 Answers: 2 Comments: 0

solve: I := ∫_0 ^( ∞) ((( tanh (x) )/x) )^( 2) dx = ? m.n.

$s o l v e :$ $I := \int_{0}^{\infty} {(\frac{t a n h (x)}{x})}^{2} d x = ?$ $m . n .$

Question Number 152863 Answers: 1 Comments: 1

∫_(−∞) ^( ∞) (1/( (√(x^2 +1)))) dx

$\int_{- \infty}^{\infty} \frac{1}{\sqrt{x^{2} + 1}} d x$

Question Number 152861 Answers: 0 Comments: 0

∫_(−∞) ^( ∞) (((ln((x^(√(x^2 +1)) +1)^2 +1))^(−ln(x^2 +1)) )/( (√(x^(∣⌊x⌋∣) +1)))) dx

$\int_{- \infty}^{\infty} \frac{{(\ln ({(x^{\sqrt{x^{2} + 1}} + 1)}^{2} + 1))}^{- \ln (x^{2} + 1)}}{\sqrt{x^{∣ ⌊ x ⌋ ∣} + 1}} d x$

Question Number 152866 Answers: 0 Comments: 0

∫_(−∞) ^( ∞) ((ln((√(x^4 +1))))/((ln(((√(x^2 +1)))^3 ))^2 )) dx

$\int_{- \infty}^{\infty} \frac{\ln (\sqrt{x^{4} + 1})}{{(\ln ({(\sqrt{x^{2} + 1})}^{3}))}^{2}} d x$

Question Number 152839 Answers: 0 Comments: 0

Prove that : Ω=∫_0 ^( 1) (( ln^( 3) (1 + x ))/x^( 2) )dx = (3/4) ζ (3 )− 2ln^( 3) ( 2 ) ■ Prepared by: M.N

$Prove that :$ $Ω = \int_{0}^{1} \frac{\ln^{3} (1 + x)}{x^{2}} d x = \frac{3}{4} ζ (3) - {2 \ln}^{3} (2) ◼$ $Prepared by : M . N$

Question Number 152778 Answers: 0 Comments: 0

Solve .......... Ω := ∫_0 ^( 1) x. sin( ln (x ))dx =^? ((−1)/( 5)) solution.... Ω :=^(i.b.p) [ (x^( 2) /2) . sin(ln(x))]_0 ^( 1) −(1/2)∫_0 ^( 1) x.cos( ln (x ))dx := ((−1)/2) ∫_0 ^( 1) x. cos (ln (x ))dx :=^(i.b.p) ((−1)/2) {[ (x^( 2) /2) cos (ln(x ))]_0 ^1 +(1/2) ∫x. sin(ln(x ))dx} := ((−1)/4) −(1/4) Ω (5/4) Ω = ((−1)/4) ⇒ Ω := ((−1)/( 5)) .........■ m.n .................................

$Solve . . . . . . . . . .$ $Ω := \int_{0}^{1} x . s i n (l n (x)) d x \overset{?}{=} \frac{- 1}{5}$ $solution . . . .$ $Ω : \overset{i . b . p}{=} {[\frac{x^{2}}{2} . s i n (l n (x))]}_{0}^{1} - \frac{1}{2} \int_{0}^{1} x . c o s (l n (x)) d x$ $:= \frac{- 1}{2} \int_{0}^{1} x . c o s (l n (x)) d x$ $: \overset{i . b . p}{=} \frac{- 1}{2} {{[\frac{x^{2}}{2} c o s (l n (x))]}_{0}^{1} + \frac{1}{2} \int x . s i n (l n (x)) d x}$ $:= \frac{- 1}{4} - \frac{1}{4} Ω$ $\frac{5}{4} Ω = \frac{- 1}{4} \Rightarrow Ω := \frac{- 1}{5} . . . . . . . . . ◼ m . n$ $. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$

Pg 58 Pg 59 Pg 60 Pg 61 Pg 62 Pg 63 Pg 64 Pg 65 Pg 66 Pg 67

Contact: info@tinkutara.com