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IntegrationQuestion and Answers: Page 92

Question Number 136840 Answers: 3 Comments: 0

∫(√(x/(x−1))) dx

$\int \sqrt{\frac{x}{x - 1}} d x$

Question Number 136841 Answers: 3 Comments: 0

∫(√(x^2 +1 )) dx = ?

$\int \sqrt{x^{2} + 1} d x = ?$

Question Number 136835 Answers: 0 Comments: 0

.....nice calculus... for f(x)=f(x+π) : ∫_(−∞) ^( ∞) f(x).((sin^2 (x))/x^2 )dx=^(why???) ∫_0 ^( π) f(x)dx

$. . . . . n i c e c a l c u l u s . . .$ $f o r f (x) = f (x + π) :$ $\int_{- \infty}^{\infty} f (x) . \frac{{s i n}^{2} (x)}{x^{2}} d x \overset{w h y ? ? ?}{=} \int_{0}^{π} f (x) d x$

Question Number 136813 Answers: 1 Comments: 0

......advanced calculus(I).... if : Ω=∫_(π/4) ^( (π/2)) ((x.cos(2x).cos(x))/(sin^7 (x))) dx=((aπ)/(90)) then :: a=???

$. . . . . . a d v a n c e d c a l c u l u s (I) . . . .$ $i f :$ $Ω = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{x . c o s (2 x) . c o s (x)}{{s i n}^{7} (x)} d x = \frac{a π}{90}$ $t h e n :: a = ? ? ?$

Question Number 136806 Answers: 2 Comments: 0

Question Number 136803 Answers: 1 Comments: 0

.....Advanced ◂.............▶ Calculus..... 𝛗=∫_(−∞) ^( ∞) ((sin((1/x^3 )))/x)dx=......??? solution:: 𝛗=^((1/x) =t) 2∫_0 ^( ∞) ((sin(t^3 ))/(1/t)).(dt/t^2 ) ............. =2∫_0 ^( ∞) ((sin(t^3 ))/t)dt ......... =^(t^3 =y) (2/3)∫_0 ^( ∞) ((sin(y))/y^(1/3) ).(dy/y^(2/3) )=(2/3)∫_0 ^( ∞) ((sin(y))/y) =^(⟨∫_0 ^( ∞) ((sin(r))/r)dr =(π/2)⟩) (π/3) ........... ........𝛗=∫_(−∞) ^( ∞) ((sin((1/x))^3 )/x)dx=(π/3) ........✓✓✓

$. . . . . A d v a n c e d ◂ . . . . . . . . . . . . . ▸ C a l c u l u s . . . . .$ $ϕ = \int_{- \infty}^{\infty} \frac{s i n (\frac{1}{x^{3}})}{x} d x = . . . . . . ? ? ?$ $s o l u t i o n ::$ $ϕ \overset{\frac{1}{x} = t}{=} 2 \int_{0}^{\infty} \frac{s i n (t^{3})}{\frac{1}{t}} . \frac{d t}{t^{2}} . . . . . . . . . . . . .$ $= 2 \int_{0}^{\infty} \frac{s i n (t^{3})}{t} d t . . . . . . . . .$ $\overset{t^{3} = y}{=} \frac{2}{3} \int_{0}^{\infty} \frac{s i n (y)}{y^{\frac{1}{3}}} . \frac{d y}{y^{\frac{2}{3}}} = \frac{2}{3} \int_{0}^{\infty} \frac{s i n (y)}{y}$ $\overset{⟨ \int_{0}^{\infty} \frac{s i n (r)}{r} d r = \frac{π}{2} ⟩}{=} \frac{π}{3} . . . . . . . . . . .$ $. . . . . . . . ϕ = \int_{- \infty}^{\infty} \frac{s i n {(\frac{1}{x})}^{3}}{x} d x = \frac{π}{3} . . . . . . . . ✓ ✓ ✓$

Question Number 136762 Answers: 0 Comments: 0

let C(a,r)={z∈C, ∣z−a∣=r } Let u,v,w ∈C(a,r) such as u+v=2w Prove that ((u−a)/(v−a)) =1 , u=v=w It shows that the middle of a segment joining two points in a circle is not in that circle

$l e t C (a, r) = {z \in C, ∣ z - a ∣= r}$ $L e t u, v, w \in C (a, r) s u c h a s u + v = 2 w$ $P r o v e t h a t \frac{u - a}{v - a} = 1, u = v = w$ $I t s h o w s t h a t t h e m i d d l e o f a s e g m e n t$ $j o i n i n g t w o p o i n t s i n a c i r c l e i s n o t i n t h a t c i r c l e$

Question Number 136750 Answers: 3 Comments: 0

.....advanced calculus..... prove that :: ... 𝛗=∫_0 ^( ∞) ((1−e^(−x^2 ) )/x^2 )dx=(√π)

$. . . . . a d v a n c e d c a l c u l u s . . . . .$ $p r o v e t h a t ::$ $. . . ϕ = \int_{0}^{\infty} \frac{1 - e^{- x^{2}}}{x^{2}} d x = \sqrt{π}$

Question Number 136723 Answers: 1 Comments: 0

....calculus (I)..... prove that :: f(x)= (1/( (√(1−4x)))) =^(???) Σ_(n=0) ^∞ ((( 2n)),(( n)) ) x^n

$. . . . c a l c u l u s (I) . . . . .$ $p r o v e t h a t ::$ $f (x) = \frac{1}{\sqrt{1 - 4 x}} \overset{? ? ?}{=} \underset{n = 0}{\sum^{\infty}} (\begin{matrix} 2 n \\ n \end{matrix}) x^{n}$

Question Number 136680 Answers: 2 Comments: 0

Question Number 136651 Answers: 2 Comments: 0

Question Number 136644 Answers: 1 Comments: 0

𝛗=∫_0 ^( 1) ((ln(x^2 +1))/x^2 )dx f(a)=∫_0 ^( 1) ((log(ax^2 +1))/x^2 )dx f ′(a)=∫_0 ^( 1) (x^2 /((ax^2 +1)x^2 ))dx=(1/a)∫_0 ^( 1) (dx/(x^2 +((1/( (√a))))^2 )) =((√a)/a)[tan^(−1) (x(√a) )]_0 ^1 =((√a)/a)tan^(−1) ((√a)) f(a)=^((√a) =u) ∫2tan^(−1) (u)du =2{u.tan^(−1) (u)−∫(u/(1+u^2 ))du}+C =2(√a) tan^(−1) ((√a) )−ln(1+a)+C f(0)=0=0+C⇒C=0 f(1)=𝛗=2((π/4))−ln(2)=(π/2)−ln(2)

$ϕ = \int_{0}^{1} \frac{l n (x^{2} + 1)}{x^{2}} d x$ $f (a) = \int_{0}^{1} \frac{l o g ({a x}^{2} + 1)}{x^{2}} d x$ $f^{'} (a) = \int_{0}^{1} \frac{x^{2}}{({a x}^{2} + 1) x^{2}} d x = \frac{1}{a} \int_{0}^{1} \frac{d x}{x^{2} + {(\frac{1}{\sqrt{a}})}^{2}}$ $= \frac{\sqrt{a}}{a} {[{t a n}^{- 1} (x \sqrt{a})]}_{0}^{1} = \frac{\sqrt{a}}{a} {t a n}^{- 1} (\sqrt{a})$ $f (a) \overset{\sqrt{a} = u}{=} \int 2 {t a n}^{- 1} (u) d u$ $= 2 {u . {t a n}^{- 1} (u) - \int \frac{u}{1 + u^{2}} d u} + C$ $= 2 \sqrt{a} {t a n}^{- 1} (\sqrt{a}) - l n (1 + a) + C$ $f (0) = 0 = 0 + C \Rightarrow C = 0$ $f (1) = ϕ = 2 (\frac{π}{4}) - l n (2) = \frac{π}{2} - l n (2)$

Question Number 136626 Answers: 1 Comments: 0

..........nice calculus......... suppose that::: ϕ(p)=∫_0 ^( ∞) ((ln(1+x))/((p+x)^2 )) ...✓ find the value of:: ∫^( 1) _0 ((ϕ(p))/(1+p))dp=?...

$. . . . . . . . . . n i c e c a l c u l u s . . . . . . . . .$ $s u p p o s e t h a t :::$ $φ (p) = \int_{0}^{\infty} \frac{l n (1 + x)}{{(p + x)}^{2}} . . . ✓$ $f i n d t h e v a l u e o f ::$ ${\int_{0}}^{1} \frac{φ (p)}{1 + p} d p = ? . . .$

Question Number 136572 Answers: 3 Comments: 1

....advanced calculus.... 𝛗=∫_0 ^( (π/2)) x.(tan(x))^(1/2) dx=??

$. . . . a d v a n c e d c a l c u l u s . . . .$ $ϕ = \int_{0}^{\frac{π}{2}} x . {(t a n (x))}^{\frac{1}{2}} d x = ? ?$

Question Number 136497 Answers: 1 Comments: 2

......nice calculus..... prove:: ∫_0 ^( 1) (1/(1+ln^2 (x)))dx=∫_(0 ) ^( ∞) ((sin(x))/(1+x))dx

$. . . . . . n i c e c a l c u l u s . . . . .$ $p r o v e ::$ $\int_{0}^{1} \frac{1}{1 + {l n}^{2} (x)} d x = \int_{0}^{\infty} \frac{s i n (x)}{1 + x} d x$

Question Number 136482 Answers: 0 Comments: 0

f(x)=∫_(−Π/4) ^(Π∫/4) e^(xtant) dt

$f (x) = \int_{- Π / 4}^{Π \int / 4} e^{x t a n t} d t$

Question Number 136481 Answers: 1 Comments: 0

∫_0 ^((50π)/3) ∣sinx∣dx

$\int_{0}^{\frac{50 π}{3}} ∣ s i n x ∣ d x$

Question Number 136476 Answers: 1 Comments: 0

(a) Let I(α)=∫_0 ^∞ e^(−(x−(α/x))^2 ) dx Show that it is legitimate to take the derivative of I(α) and also I′(α)= 0. Then show that I(α)=((√π)/2). (b) Use (a) to prove ∫_0 ^∞ e^(−(x^2 +α^2 x^(−2) )) dx=((√π)/2)e^(−2α) .

$(a) Let$ $I (α) = \int_{0}^{\infty} e^{- {(x - \frac{α}{x})}^{2}} dx$ $Show that it is legitimate to take the derivative of I (α) and also I^{'} (α) =$ $0 . Then show that$ $I (α) = \frac{\sqrt{π}}{2} .$ $(b) Use (a) to prove$ $\int_{0}^{\infty} e^{- (x^{2} + α^{2} x^{- 2})} dx = \frac{\sqrt{π}}{2} e^{- 2 α} .$