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IntegrationQuestion and Answers: Page 36

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find the value of b so that the line y=b divides the region bound by the graphs of the two functinos , into two regions of equal area. f(x)=9−x^2 and g(x)=0

$f i n d t h e v a l u e o f b s o t h a t t h e l i n e y = b$ $d i v i d e s t h e r e g i o n b o u n d b y t h e g r a p h s o f$ $t h e t w o f u n c t i n o s, i n t o t w o r e g i o n s o f e q u a l$ $a r e a .$ $f (x) = 9 - x^{2} a n d g (x) = 0$

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𝛗 = ∫_0 ^( ∞) ((cos (ax)− sin(bx))/x^( 2) )dx =? 𝛗=∫_0 ^( ∞) ((1−2sin^( 2) (((ax)/2))−(1−2sin^( 2) (((bx)/2))))/(x^2 ))dx =2 {∫_0 ^( ∞) (((sin(((bx)/2)))/x))^2 dx=Θ_1 }−2{∫_0 ^( ∞) (((sin(((ax)/2)))/x))^2 dx=Θ_2 } Θ_( 1) =^(((bx)/2)=t) (b/2)∫_0 ^( ∞) ((sin^( 2) (t))/t^( 2) )dt= ((πb)/4) similarly : Θ_( 2) =((πa)/4) ∴ 𝛗= (π/2)(∣b∣−∣a∣) Dirichlet′s integrals: ∫_0 ^( ∞) ((sin(x))/x)dx=(π/2)=∫_0 ^( ∞) ((sin^2 (x))/x^( 2) )dx

$ϕ = \int_{0}^{\infty} \frac{c o s (a x) - s i n (b x)}{x^{2}} d x = ?$ $ϕ = \int_{0}^{\infty} \frac{1 - 2 {s i n}^{2} (\frac{a x}{2}) - (1 - 2 {s i n}^{2} (\frac{b x}{2}))}{x^{2}} d x$ $= 2 {\int_{0}^{\infty} {(\frac{s i n (\frac{b x}{2})}{x})}^{2} d x = Θ_{1}} - 2 {\int_{0}^{\infty} {(\frac{s i n (\frac{a x}{2})}{x})}^{2} d x = Θ_{2}}$ $Θ_{1} \overset{\frac{b x}{2} = t}{=} \frac{b}{2} \int_{0}^{\infty} \frac{{s i n}^{2} (t)}{t^{2}} d t = \frac{π b}{4}$ $s i m i l a r l y : Θ_{2} = \frac{π a}{4} ∴ ϕ = \frac{π}{2} (∣ b ∣ - ∣ a ∣)$ ${D i r i c h l e t}^{'} s i n t e g r a l s : \int_{0}^{\infty} \frac{s i n (x)}{x} d x = \frac{π}{2} = \int_{0}^{\infty} \frac{{s i n}^{2} (x)}{x^{2}} d x$

Question Number 173424 Answers: 1 Comments: 0

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

$prove that$ $\int_{0}^{\infty} \frac{\sin (x)}{\sinh (x)} dx = \frac{π}{2} \tanh (\frac{π}{2})$

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let ∀n∈N: I_n (f(x))= the n^(th) antiderivate of f(x) with I_0 =f(x) find the formula for the constants a_n , b_n of I_n (ln x)=a_n x^n ln x +b_n x^n

$let \forall n \in N : I_{n} (f (x)) = the n^{th} antiderivate$ $of f (x) with I_{0} = f (x)$ $find the formula for the constants a_{n}, b_{n} of$ $I_{n} (\ln x) = a_{n} x^{n} \ln x + b_{n} x^{n}$

Question Number 173149 Answers: 0 Comments: 1

∫_0 ^( ∞) (( dx)/( (√(1+x)) .(2+2x +x^( 2) )))=(1/σ) (π−ln(3+2(√3) )) σ = ? −− solution −− Ω=^((√(1+x)) =t) 2∫_0 ^( ∞) (dt/( 1+ t^( 4) )) = 2∫_0 ^( 1) (dt/(1 + t^( 4) )) Ψ = ∫_0 ^( 1) (( dt)/(1+t^( 4) )) = (1/2)∫_0 ^( 1) (( 1+t^( 2) −(t^( −2) −1))/(1+t^( 4) ))dt = (1/2) ∫_0 ^( 1) (( 1+ t^( 2) )/(1+t^( 4) )) dt +(1/2) ∫_0 ^( 1) ((1−t^( 2) )/(1+ t^( 4) )) dt Φ=∫_0 ^( 1) ((1 +t^( 2) )/(1+t^( 4) ))dt =∫_0 ^( 1) (( t^( −2) +1)/(t^( −2) + t^( 2) )) dt =∫_0 ^( 1) (( 1+t^( −2) )/(( t^ − t^( −1) )^( 2) +2)) =^(sub) [ (1/( (√2))) tan^( −1) (t −t^( −1) )]_0 ^1 =(π/(2(√2))) ∗ 𝛗 = ∫_0 ^( 1) (( 1−t^( 2) )/(1+t^( 4) )) dt = ∫_0 ^( 1) ((t^( −2) −1)/(( t +t^( −1) )^( 2) −2))dt =^(t +(1/t) =u) −∫_2 ^( ∞) (du/(( u−(√2) )(u+ (√2) ))) = ((−1)/(2(√2))) [ln(((u −(√2))/(u+(√2))))]_2 ^( ∞) =(1/(2(√2))) ln(((2−(√2))/(2+(√2))) ) 𝛗 =−(1/(2(√2))) ln( 3 +2(√2) ) ∗∗ (∗) & (∗∗):: Ω=2Ψ= (Φ + 𝛗) =(1/(2(√2))) ( π −ln( 3 +2(√2) ) ...■m.n

$\int_{0}^{\infty} \frac{d x}{\sqrt{1 + x} . (2 + 2 x + x^{2})} = \frac{1}{σ} (π - \ln (3 + 2 \sqrt{3}))$ $σ = ?$ $- - solution - -$ $Ω \overset{\sqrt{1 + x} = t}{=} 2 \int_{0}^{\infty} \frac{d t}{1 + t^{4}} = 2 \int_{0}^{1} \frac{d t}{1 + t^{4}}$ $Ψ = \int_{0}^{1} \frac{d t}{1 + t^{4}} = \frac{1}{2} \int_{0}^{1} \frac{1 + t^{2} - (t^{- 2} - 1)}{1 + t^{4}} d t$ $= \frac{1}{2} \int_{0}^{1} \frac{1 + t^{2}}{1 + t^{4}} d t + \frac{1}{2} \int_{0}^{1} \frac{1 - t^{2}}{1 + t^{4}} d t$ $Φ = \int_{0}^{1} \frac{1 + t^{2}}{1 + t^{4}} d t = \int_{0}^{1} \frac{t^{- 2} + 1}{t^{- 2} + t^{2}} d t$ $= \int_{0}^{1} \frac{1 + t^{- 2}}{{(t^{} - t^{- 1})}^{2} + 2} \overset{s u b}{=} {[\frac{1}{\sqrt{2}} {t a n}^{- 1} (t - t^{- 1})]}_{0}^{1} = \frac{π}{2 \sqrt{2}} *$ $ϕ = \int_{0}^{1} \frac{1 - t^{2}}{1 + t^{4}} d t = \int_{0}^{1} \frac{t^{- 2} - 1}{{(t + t^{- 1})}^{2} - 2} d t$ $\overset{t + \frac{1}{t} = u}{=} - \int_{2}^{\infty} \frac{d u}{(u - \sqrt{2}) (u + \sqrt{2})}$ $= \frac{- 1}{2 \sqrt{2}} {[\ln (\frac{u - \sqrt{2}}{u + \sqrt{2}})]}_{2}^{\infty} = \frac{1}{2 \sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})$ $ϕ = - \frac{1}{2 \sqrt{2}} \ln (3 + 2 \sqrt{2}) * *$ $(*) & (* *) :: Ω = 2 Ψ = (Φ + ϕ) = \frac{1}{2 \sqrt{2}} (π - \ln (3 + 2 \sqrt{2}) . . . ◼ m . n$

Question Number 173148 Answers: 1 Comments: 0

∫_0 ^1 ^n (√x) (arcsin x) dx

$\int_{0}^{1}^{n} \sqrt{x} (\arcsin x) dx$

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let U_n =∫_0 ^1 (√(1−x^n ))ln^2 xdx 1)lim U_n ? 2)equivalent of U_n (n→∞)

$l e t U_{n} = \int_{0}^{1} \sqrt{1 - x^{n}} {l n}^{2} x d x$ $1) l i m U_{n} ?$ $2) e q u i v a l e n t o f U_{n} (n \to \infty)$

Question Number 181403 Answers: 1 Comments: 0

Refer to Q181319 ∫((x^4 +1)/(x^4 +x+2))dx ((x^4 +1)/(x^4 +x+2))=1+(x/(2(x^2 +x+1)))−(x/(2(x^2 −x+1))) =1+(1/4)[(((2x+1)−1)/(x^2 +x+1))−(1/4)×(((2x−1)+1)/((x^2 −x+1))) =(1/4)×[log(x^2 +x+1)]^′ −(1/4)×((1/([((√3)/2)(x+(1/2))^2 ]+1))) −((1/4)[log(x^2 −x+1)]^′ +(1/4)((1/([((√3)/2)×(x−(1/2)]^2 +1)))) first with [((√3)/2)( x+(1/2))]=u ∫(1/(u^2 +1))du=arctg(u) and v=[((√3)/2)(x−(1/2) )]=v ∫(dv/(1+v^2 ))=arctg(v) ...............

$R e f e r t o Q 181319$ $\int \frac{x^{4} + 1}{x^{4} + x + 2} d x$ $\frac{x^{4} + 1}{x^{4} + x + 2} = 1 + \frac{x}{2 (x^{2} + x + 1)} - \frac{x}{2 (x^{2} - x + 1)}$ $= 1 + \frac{1}{4} [\frac{(2 x + 1) - 1}{x^{2} + x + 1} - \frac{1}{4} \times \frac{(2 x - 1) + 1}{(x^{2} - x + 1)}$ $= \frac{1}{4} \times {[\log (x^{2} + x + 1)]}^{^{'}} - \frac{1}{4} \times (\frac{1}{[\frac{\sqrt{3}}{2} {(x + \frac{1}{2})}^{2}] + 1})$ $- (\frac{1}{4} {[\log (x^{2} - x + 1)]}^{^{'}} + \frac{1}{4} (\frac{1}{[\frac{\sqrt{3}}{2} \times {(x - \frac{1}{2}]}^{2} + 1}))$ $f i r s t w i t h [\frac{\sqrt{3}}{2} (x + \frac{1}{2})] = u \int \frac{1}{u^{2} + 1} d u = a r c t g (u)$ $a n d v = [\frac{\sqrt{3}}{2} (x - \frac{1}{2})] = v \int \frac{d v}{1 + v^{2}} = a r c t g (v)$ $. . . . . . . . . . . . . . .$