Question and Answers Forum

IntegrationQuestion and Answers: Page 100

Question Number 136858 Answers: 1 Comments: 0

calculate ∫_0 ^∞ ((cos(sinx)−sin(cosx))/((x^2 +1)^2 ))dx

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{sinx}\right)−\mathrm{sin}\left(\mathrm{cosx}\right)}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 136844 Answers: 0 Comments: 0

Question Number 136840 Answers: 3 Comments: 0

∫(√(x/(x−1))) dx

$$\:\int\sqrt{\frac{{x}}{{x}−\mathrm{1}}}\:{dx} \\ $$

Question Number 136841 Answers: 3 Comments: 0

∫(√(x^2 +1 )) dx = ?

$$\int\sqrt{{x}^{\mathrm{2}} +\mathrm{1}\:}\:{dx}\:=\:? \\ $$

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

$$\:\:\:\:\:\:\:\:.....{nice}\:\:\:\:\:{calculus}... \\ $$$$\:\:\:{for}\:\:\:\:{f}\left({x}\right)={f}\left({x}+\pi\right)\:: \\ $$$$\:\:\:\:\:\int_{−\infty} ^{\:\infty} {f}\left({x}\right).\frac{{sin}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}\overset{{why}???} {=}\int_{\mathrm{0}} ^{\:\pi} {f}\left({x}\right){dx} \\ $$

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=???

$$\:\:\:\:\: \\ $$$$\:\:\:......{advanced}\:\:\:\:{calculus}\left({I}\right).... \\ $$$$\:\:\:\:{if}\::\: \\ $$$$\:\:\Omega=\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{x}.{cos}\left(\mathrm{2}{x}\right).{cos}\left({x}\right)}{{sin}^{\mathrm{7}} \left({x}\right)}\:{dx}=\frac{{a}\pi}{\mathrm{90}} \\ $$$$\:\:\:{then}\:\:::\:\:{a}=??? \\ $$$$\:\: \\ $$$$\:\:\:\: \\ $$

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) ........✓✓✓

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:.....\mathscr{A}{dvanced}\:\:\:\:\blacktriangleleft.............\blacktriangleright\:\:\:\mathscr{C}{alculus}..... \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)}{{x}}{dx}=......??? \\ $$$$\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{\frac{\mathrm{1}}{{x}}\:={t}} {=}\:\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{\frac{\mathrm{1}}{{t}}}.\frac{{dt}}{{t}^{\mathrm{2}} }\:............. \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({t}^{\mathrm{3}} \right)}{{t}}{dt}\:......... \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{{t}^{\mathrm{3}} ={y}} {=}\:\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}^{\frac{\mathrm{1}}{\mathrm{3}}} }.\frac{{dy}}{{y}^{\frac{\mathrm{2}}{\mathrm{3}}} }=\frac{\mathrm{2}}{\mathrm{3}}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({y}\right)}{{y}} \\ $$$$\:\:\:\:\overset{\langle\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({r}\right)}{{r}}{dr}\:=\frac{\pi}{\mathrm{2}}\rangle} {=}\:\frac{\pi}{\mathrm{3}}\:........... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:........\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \frac{{sin}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} }{{x}}{dx}=\frac{\pi}{\mathrm{3}}\:\:........\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

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

$$\:{let}\:\:{C}\left({a},{r}\right)=\left\{{z}\in\mathbb{C},\:\:\mid{z}−{a}\mid={r}\:\right\} \\ $$$${Let}\:{u},{v},{w}\:\in{C}\left({a},{r}\right)\:{such}\:{as}\:\:\:{u}+{v}=\mathrm{2}{w} \\ $$$${Prove}\:{that}\:\:\frac{{u}−{a}}{{v}−{a}}\:=\mathrm{1}\:,\:{u}={v}={w} \\ $$$$ \\ $$$${It}\:{shows}\:{that}\:{the}\:{middle}\:{of}\:{a}\:{segment}\: \\ $$$${joining}\:{two}\:{points}\:{in}\:{a}\:{circle}\:\:{is}\:{not}\:{in}\:{that}\:{circle} \\ $$

Question Number 136750 Answers: 3 Comments: 0

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

$$\:\:\:\:\:\:\:\:\:\:\:.....{advanced}\:\:\:\:{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:...\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{e}^{−{x}^{\mathrm{2}} } }{{x}^{\mathrm{2}} }{dx}=\sqrt{\pi}\:\:\:\: \\ $$

Question Number 136723 Answers: 1 Comments: 0

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

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{calculus}\:\:\:\:\:\left({I}\right)..... \\ $$$$\:\:\:\:\:{prove}\:\:\:{that}\::: \\ $$$$\:\:\:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}\:\overset{???} {=}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\:\mathrm{2}{n}}\\{\:\:\:{n}}\end{pmatrix}\:{x}^{{n}} \\ $$$$ \\ $$

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)

$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{log}\left({ax}^{\mathrm{2}} +\mathrm{1}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:{f}\:'\left({a}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({ax}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{{x}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\sqrt{{a}}}{{a}}\left[{tan}^{−\mathrm{1}} \left({x}\sqrt{{a}}\:\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\sqrt{{a}}}{{a}}{tan}^{−\mathrm{1}} \left(\sqrt{{a}}\right) \\ $$$$\:\:{f}\left({a}\right)\overset{\sqrt{{a}}\:={u}} {=}\int\mathrm{2}{tan}^{−\mathrm{1}} \left({u}\right){du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left\{{u}.{tan}^{−\mathrm{1}} \:\left({u}\right)−\int\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\right\}+{C} \\ $$$$\:=\mathrm{2}\sqrt{{a}}\:{tan}^{−\mathrm{1}} \left(\sqrt{{a}}\:\right)−{ln}\left(\mathrm{1}+{a}\right)+{C} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}=\mathrm{0}+{C}\Rightarrow{C}=\mathrm{0} \\ $$$$\:\:\:{f}\left(\mathrm{1}\right)=\boldsymbol{\phi}=\mathrm{2}\left(\frac{\pi}{\mathrm{4}}\right)−{ln}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{2}}−{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

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=?...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:..........{nice}\:\:\:\:{calculus}......... \\ $$$$\:\:\:\:{suppose}\:{that}::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\varphi\left({p}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({p}+{x}\right)^{\mathrm{2}} }\:...\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:{find}\:{the}\:{value}\:{of}:: \\ $$$$\:\:\:\:\:\:\:\underset{\mathrm{0}} {\int}^{\:\mathrm{1}} \:\frac{\varphi\left({p}\right)}{\mathrm{1}+{p}}{dp}=?... \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 136572 Answers: 3 Comments: 1

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

$$\:\:\:\:\:\:\:\:\:\:\:\:\:....{advanced}\:\:\:\:{calculus}.... \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {x}.\left({tan}\left({x}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} {dx}=?? \\ $$$$ \\ $$

Question Number 136497 Answers: 1 Comments: 2

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

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......{nice}\:\:\:\:{calculus}..... \\ $$$$\:\:\:\:{prove}::\:\: \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{ln}^{\mathrm{2}} \left({x}\right)}{dx}=\int_{\mathrm{0}\:} ^{\:\infty} \frac{{sin}\left({x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$ \\ $$

Question Number 136482 Answers: 0 Comments: 0

f(x)=∫_(−Π/4) ^(Π∫/4) e^(xtant) dt

$${f}\left({x}\right)=\int_{−\Pi/\mathrm{4}} ^{\Pi\int/\mathrm{4}} {e}^{{xtant}} {dt} \\ $$

Question Number 136481 Answers: 1 Comments: 0

∫_0 ^((50π)/3) ∣sinx∣dx

$$\int_{\mathrm{0}} ^{\frac{\mathrm{50}\pi}{\mathrm{3}}} \mid{sinx}\mid{dx} \\ $$

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) .

Question Number 136473 Answers: 0 Comments: 0

Question Number 136445 Answers: 2 Comments: 0

Question Number 136440 Answers: 2 Comments: 0

∫ (dx/(sin^6 x)) ?

$$\:\int\:\frac{{dx}}{\mathrm{sin}\:^{\mathrm{6}} {x}}\:? \\ $$

Question Number 136425 Answers: 1 Comments: 3

If α>0 and β>0, prove ∫_0 ^∞ ((ln(αx))/(β^2 +x^2 ))dx=(π/(2β))ln(αβ)

$$\mathrm{If}\:\alpha>\mathrm{0}\:\mathrm{and}\:\beta>\mathrm{0},\:\mathrm{prove} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\alpha\mathrm{x}\right)}{\beta^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dx}=\frac{\pi}{\mathrm{2}\beta}\mathrm{ln}\left(\alpha\beta\right) \\ $$

Question Number 136406 Answers: 0 Comments: 1

calculate A_λ =∫_0 ^∞ ((cos^4 x)/((x^2 +λ^2 )^2 ))dx 2) find the value of ∫_0 ^∞ ((cos^4 x)/((x^2 +3)^2 ))dx

$$\mathrm{calculate}\:\:\mathrm{A}_{\lambda} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}^{\mathrm{4}} \mathrm{x}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 136405 Answers: 0 Comments: 0

if f(x)=x^3 −3x+2 determine f^(−1) (x) and ∫ f^(−1) (nf(x))dx with n integr

$$\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} −\mathrm{3x}+\mathrm{2}\:\:\mathrm{determine}\:\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{and}\:\int\:\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{nf}\left(\mathrm{x}\right)\right)\mathrm{dx}\:\:\mathrm{with} \\ $$$$\mathrm{n}\:\mathrm{integr} \\ $$

Pg 95 Pg 96 Pg 97 Pg 98 Pg 99 Pg 100 Pg 101 Pg 102 Pg 103 Pg 104

Contact: info@tinkutara.com