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Question Number 54074    Answers: 1   Comments: 5

Evaluate : 1) ∫_0 ^( 1) (dx/((√(1+x))+(√(1−x))+2)) 2) ∫_0 ^( 2) ((ln(1+2x))/(1+x^2 )) 3) ∫_0 ^( π) (x/(√(1+sin^3 x)))((3πcosx+4sinx)sin^2 x+4)dx 4) ∫_0 ^( π) ((x^2 cos^2 x−xsinx−cosx−1)/((1+xsinx)^2 )) dx.

$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right)\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}+\mathrm{2}}\: \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\mathrm{2}{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right) \\ $$$$\:\int_{\mathrm{0}} ^{\:\pi} \frac{{x}}{\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} {x}}}\left(\left(\mathrm{3}\pi\mathrm{cos}{x}+\mathrm{4sin}{x}\right)\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{4}\right){dx} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:\frac{{x}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}−{x}\mathrm{sin}{x}−\mathrm{cos}{x}−\mathrm{1}}{\left(\mathrm{1}+{x}\mathrm{sin}{x}\right)^{\mathrm{2}} }\:{dx}. \\ $$

Question Number 54073    Answers: 1   Comments: 0

Question Number 54070    Answers: 2   Comments: 7

Evaluate : 1) ∫_(−1) ^( 1) cot^(−1) ((1/(√(1−x^2 )))).(cot^(−1) (x/(√(1−(x^2 )^(∣x∣) ))))dx 2) ∫_0 ^( (π/2)) ((sin^2 (10)θ)/(sin^2 θ)) dθ 3) ∫_0 ^( (π/4)) ((ln(cotx))/(((sinx)^(2009) +(cosx)^(2009) )^2 )).(sin2x)^(2008) dx 4) ∫_0 ^( 2) ((4x+10)/((x^2 +5x+6)^2 )) dx.

$${Evaluate}\:: \\ $$$$\left.\mathrm{1}\right) \\ $$$$\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \mathrm{cot}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right).\left(\mathrm{co}{t}^{−\mathrm{1}} \frac{{x}}{\sqrt{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{\mid{x}\mid} }}\right){dx} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{10}\right)\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$\left.\mathrm{3}\right) \\ $$$$\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left({cotx}\right)}{\left(\left(\mathrm{sin}{x}\right)^{\mathrm{2009}} +\left(\mathrm{cos}{x}\right)^{\mathrm{2009}} \right)^{\mathrm{2}} }.\left(\mathrm{sin2}{x}\right)^{\mathrm{2008}} {dx} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \:\frac{\mathrm{4}{x}+\mathrm{10}}{\left({x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{6}\right)^{\mathrm{2}} }\:{dx}. \\ $$

Question Number 54068    Answers: 1   Comments: 2

Question Number 54061    Answers: 0   Comments: 8

Question Number 54051    Answers: 1   Comments: 0

Question Number 54045    Answers: 2   Comments: 1

Question Number 54034    Answers: 0   Comments: 1

If f(x)=f(a+b−x), then ∫_a ^b x f(x) dx =

$$\mathrm{If}\:{f}\left({x}\right)={f}\left({a}+{b}−{x}\right),\:\mathrm{then}\:\underset{{a}} {\overset{{b}} {\int}}\:{x}\:{f}\left({x}\right)\:{dx}\:= \\ $$

Question Number 54033    Answers: 2   Comments: 1

∫_( 0) ^1 (√((1−x)/(1+x))) dx =

$$\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:{dx}\:= \\ $$

Question Number 54032    Answers: 0   Comments: 4

∫_(−2 ) ^2 min (x−[x], −x−[−x])dx equals [x] represents greatest integer less than or equal to x).

$$\underset{−\mathrm{2}\:} {\overset{\mathrm{2}} {\int}}\:\mathrm{min}\:\left({x}−\left[{x}\right],\:−{x}−\left[−{x}\right]\right){dx}\:\mathrm{equals}\:\left[{x}\right] \\ $$$$\mathrm{represents}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{less}\:\mathrm{than}\:\mathrm{or} \\ $$$$\left.\mathrm{equal}\:\mathrm{to}\:{x}\right). \\ $$

Question Number 54031    Answers: 1   Comments: 0

If f(a+b−x)= f(x), then ∫_a ^b x f(x) dx =

$$\mathrm{If}\:\:{f}\left({a}+{b}−{x}\right)=\:{f}\left({x}\right),\:\mathrm{then}\:\underset{{a}} {\overset{{b}} {\int}}\:{x}\:{f}\left({x}\right)\:{dx}\:= \\ $$

Question Number 54029    Answers: 1   Comments: 1

∫_( 3) ^9 ((√x)/((√(12−x)) + (√x))) dx = 9

$$\underset{\:\mathrm{3}} {\overset{\mathrm{9}} {\int}}\:\frac{\sqrt{{x}}}{\sqrt{\mathrm{12}−{x}}\:+\:\sqrt{{x}}}\:{dx}\:=\:\mathrm{9} \\ $$

Question Number 54028    Answers: 1   Comments: 0

If for every integer n, ∫_n ^(n+1) f(x) dx = n^2 , then the value of ∫_(−2) ^4 f(x) dx=

$$\mathrm{If}\:\mathrm{for}\:\mathrm{every}\:\mathrm{integer}\:{n},\:\underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{f}\left({x}\right)\:{dx}\:=\:{n}^{\mathrm{2}} , \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\underset{−\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}\right)\:{dx}= \\ $$

Question Number 54025    Answers: 1   Comments: 1

The points A, B, C, D have coordinates (−7,9), (3,4), (1,12), and (−2,−9). find the length of the linePQ where P devides AB in the ratio 2:3 and devides CD in the ratio 1:−4.

$$\mathrm{The}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C},\:\mathrm{D}\:\mathrm{have}\:\mathrm{coordinates} \\ $$$$\left(−\mathrm{7},\mathrm{9}\right),\:\left(\mathrm{3},\mathrm{4}\right),\:\left(\mathrm{1},\mathrm{12}\right),\:\mathrm{and}\:\left(−\mathrm{2},−\mathrm{9}\right). \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{linePQ}\:\mathrm{where}\:\mathrm{P} \\ $$$$\mathrm{devides}\:\mathrm{AB}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{2}:\mathrm{3}\:\mathrm{and}\:\mathrm{devides} \\ $$$$\mathrm{CD}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:−\mathrm{4}. \\ $$

Question Number 54024    Answers: 1   Comments: 0

The coefficient of x^4 in (2−4x+3x^2 )^(−2) i s ?

$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{4}} \:\mathrm{in}\:\left(\mathrm{2}−\mathrm{4x}+\mathrm{3x}^{\mathrm{2}} \right)^{−\mathrm{2}} \mathrm{i}\:\mathrm{s}\:? \\ $$

Question Number 54022    Answers: 0   Comments: 4

Question Number 54015    Answers: 1   Comments: 0

A stone is thrown vertically upwards from the top of a tower 50.0m high with an initial velocity of 20.0ms^(−1) .calculate i.the maximum height the stone reaches ii.the time it takes to reach the maximum height iii.the total distance it covers[take g=10ms^(−1) ]

$$\mathrm{A}\:\mathrm{stone}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards}\:\mathrm{from}\:\mathrm{the}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{tower}\:\mathrm{50}.\mathrm{0m}\:\mathrm{high}\:\mathrm{with}\:\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{20}.\mathrm{0ms}^{−\mathrm{1}} .\mathrm{calculate} \\ $$$$\mathrm{i}.\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{the}\:\mathrm{stone}\:\mathrm{reaches} \\ $$$$\mathrm{ii}.\mathrm{the}\:\mathrm{time}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{iii}.\mathrm{the}\:\mathrm{total}\:\mathrm{distance}\:\mathrm{it}\:\mathrm{covers}\left[\mathrm{take}\:\mathrm{g}=\mathrm{10ms}^{−\mathrm{1}} \right] \\ $$

Question Number 54013    Answers: 1   Comments: 0

2^x =−4.solve for x

$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} =−\mathrm{4}.\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}} \\ $$

Question Number 54011    Answers: 1   Comments: 2

calculate f(a) =∫ (dx/((√(1+ax))−(√(1−ax)))) with a>0 . 2) calculate U_n =∫_(−(1/(na))) ^(1/(na)) (dx/((√(1+ax))−(√(1−ax)))) with n from N and n>1 find lim_(n→+∞) U_n and study the convergence of Σ U_n

$${calculate}\:{f}\left({a}\right)\:=\int\:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+{ax}}−\sqrt{\mathrm{1}−{ax}}}\:\:{with}\:{a}>\mathrm{0}\:. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:{U}_{{n}} =\int_{−\frac{\mathrm{1}}{{na}}} ^{\frac{\mathrm{1}}{{na}}} \:\:\frac{{dx}}{\sqrt{\mathrm{1}+{ax}}−\sqrt{\mathrm{1}−{ax}}}\:\:{with}\:{n}\:{from}\:{N}\:{and}\:{n}>\mathrm{1} \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:{U}_{{n}} \:\:\:{and}\:{study}\:{the}\:{convergence}\:{of}\:\Sigma\:{U}_{{n}} \\ $$

Question Number 53998    Answers: 0   Comments: 1

In the next link, Barry Barrish, who won the physic nobel prize in 2017; gave an exclusive interview to IFT-Instituto de Fisica Teo^ rica-(part 1):

$$\mathrm{In}\:\mathrm{the}\:\mathrm{next}\:\mathrm{link},\:\mathrm{Barry}\:\mathrm{Barrish},\:\mathrm{who}\:\mathrm{won}\:\mathrm{the}\:\mathrm{physic}\:\mathrm{nobel}\:\mathrm{prize}\:\mathrm{in}\:\mathrm{2017}; \\ $$$$\mathrm{gave}\:\mathrm{an}\:\mathrm{exclusive}\:\mathrm{interview}\:\mathrm{to}\:\mathrm{IFT}-{Instituto}\:{de}\:{Fisica}\:{Te}\acute {{o}rica}-\left(\mathrm{part}\:\mathrm{1}\right): \\ $$

Question Number 53967    Answers: 2   Comments: 1

1)calculate A_t =∫_0 ^∞ e^(−xt) sinxdx with x>0 2) by using Fubuni theorem find the value of ∫_0 ^∞ ((sinx)/x)dx .

$$\left.\mathrm{1}\right){calculate}\:{A}_{{t}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} \:{sinxdx}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{by}\:{using}\:{Fubuni}\:{theorem}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}{dx}\:. \\ $$

Question Number 53966    Answers: 0   Comments: 1

let f(x) =xsinx ,2π periodic even developp f at Fourier serie .

$${let}\:{f}\left({x}\right)\:={xsinx}\:,\mathrm{2}\pi\:{periodic}\:{even} \\ $$$${developp}\:{f}\:{at}\:{Fourier}\:{serie}\:. \\ $$

Question Number 53965    Answers: 1   Comments: 0

solve (x+1)y^(′′) +(2−3x^2 )y^′ =xsin(x) let y^′ =z so (e) ⇔(x+1)z^′ +(2−3x^2 )z =xsinx(e) let first find z (he) →(x+1)z^′ +(2−3x^2 )z =0 ⇒(x+1)z^′ =(3x^2 −2)z ⇒(z^′ /z) =((3x^2 −2)/(x+1)) ⇒ ln∣z∣ = ∫ ((3x^2 −2)/(x+1)) dx +c =∫ ((3(x^2 −1)+1)/(x+1))+c =∫3(x−1)dx +∫ (dx/(x+1)) +c =3((x^2 /2)−x)+ln∣x+1∣ +c ⇒z =K∣x+1∣e^((3/2)x^2 −3x) solution on]−1,+∞[ ⇒ z(x) =K(x+1) e^((3/2)x^2 −3x) mvc method give z^′ =K^′ (x+1)e^((3/2)x^2 −3x) +K{ e^((3/2)x^2 −3x) +(x+1)(3x−3) e^((3/2)x^2 −3x) } ={ K^′ (x+1) + K +3K(x+1)(x−1) } e^((3/2)x^2 −3x) (e) ⇒{(x+1)^2 K^′ +K(x+1) +3K(x+1)^2 (x−1)} e^((3/2)x^2 −3x) (2−3x^2 )K(x+1)e^((3/2)x^2 −3x) =xsinx ⇒ (x+1)K^′ +K +3K(x^2 −1) +K(2−3x^2 ) =((xsinx)/(x+1)) ⇒ (x+1)K^′ +K −K =((xsinx)/(x+1)) ⇒K^′ =((xsinx)/((x+1)^2 )) ⇒ K(x) =∫ ((xsinx)/((x+1)^2 ))dx +c_0 ⇒z(x) =(x+1)e^((3/2)x^2 −3x) { ∫ ((xsinx)/((x+1)^2 ))dx +c_0 } =c_0 (x+1) e^((3/2)x^2 −3x) +(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 )) dt we have y^′ (x)=z(x) ⇒y(x) =∫ z(x)dx +λ ⇒ y(x) =∫ c_0 (x+1)e^((3/2)x^2 −3x) dx +∫ {(x+1)e^((3/2)x^2 −3x) ∫_. ^x ((tsint)/((t+1)^2 ))dt}dx +λ .

$${solve}\:\:\left({x}+\mathrm{1}\right){y}^{''} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){y}^{'} \:={xsin}\left({x}\right) \\ $$$${let}\:{y}^{'} ={z}\:\:\:{so}\:\left({e}\right)\:\Leftrightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){z}\:={xsinx}\left({e}\right)\:{let}\:{first}\:{find}\:{z} \\ $$$$\left({he}\right)\:\rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:+\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){z}\:=\mathrm{0}\:\Rightarrow\left({x}+\mathrm{1}\right){z}^{'} \:=\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\right){z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\:\int\:\:\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}}{{x}+\mathrm{1}}\:{dx}\:+{c}\:\:=\int\:\frac{\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}}{{x}+\mathrm{1}}+{c}\:=\int\mathrm{3}\left({x}−\mathrm{1}\right){dx}\:+\int\:\frac{{dx}}{{x}+\mathrm{1}}\:+{c} \\ $$$$\left.=\mathrm{3}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)+{ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow{z}\:={K}\mid{x}+\mathrm{1}\mid{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:{solution}\:{on}\right]−\mathrm{1},+\infty\left[\:\Rightarrow\right. \\ $$$${z}\left({x}\right)\:={K}\left({x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:\:{mvc}\:{method}\:{give} \\ $$$${z}^{'} ={K}^{'} \left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+{K}\left\{\:\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+\left({x}+\mathrm{1}\right)\left(\mathrm{3}{x}−\mathrm{3}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \right\} \\ $$$$=\left\{\:{K}^{'} \left({x}+\mathrm{1}\right)\:+\:{K}\:+\mathrm{3}{K}\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\:\right\}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \\ $$$$\left({e}\right)\:\Rightarrow\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} {K}^{'} \:+{K}\left({x}+\mathrm{1}\right)\:+\mathrm{3}{K}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}−\mathrm{1}\right)\right\}\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \\ $$$$\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right){K}\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:={xsinx}\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){K}^{'} \:+{K}\:+\mathrm{3}{K}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\:+{K}\left(\mathrm{2}−\mathrm{3}{x}^{\mathrm{2}} \right)\:=\frac{{xsinx}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\left({x}+\mathrm{1}\right){K}^{'} \:+{K}\:−{K}\:=\frac{{xsinx}}{{x}+\mathrm{1}}\:\Rightarrow{K}^{'} \:=\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int\:\:\:\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:+{c}_{\mathrm{0}} \:\Rightarrow{z}\left({x}\right)\:=\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \left\{\:\int\:\:\frac{{xsinx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:+{c}_{\mathrm{0}} \right\} \\ $$$$={c}_{\mathrm{0}} \left({x}+\mathrm{1}\right)\:{e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\:+\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\int_{.} ^{{x}} \:\:\:\frac{{tsint}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\ $$$${we}\:{have}\:{y}^{'} \left({x}\right)={z}\left({x}\right)\:\Rightarrow{y}\left({x}\right)\:=\int\:{z}\left({x}\right){dx}\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\int\:{c}_{\mathrm{0}} \left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} {dx}\:+\int\:\:\left\{\left({x}+\mathrm{1}\right){e}^{\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{3}{x}} \:\int_{.} ^{{x}} \:\:\frac{{tsint}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right\}{dx}\:+\lambda\:. \\ $$

Question Number 53956    Answers: 0   Comments: 1

give∫_0 ^1 e^(−x) ln(1−x)dx at form of serie

$${give}\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} {ln}\left(\mathrm{1}−{x}\right){dx}\:\:{at}\:{form}\:{of}\:{serie} \\ $$

Question Number 53957    Answers: 0   Comments: 1

let f(x) =arctan(1+2x) 1) calculate f^((n)) (x) then f^((n)) (0) 2) developp f at integr serie . we have f^′ (x)=(2/(1+(1+2x)^2 )) ⇒ f^((n)) (x) =2 {(1/((2x+1)^2 +1))}^((n−1)) with n>0 let W(x)=(1/((2x+1)^2 +1)) ⇒W(x) =(1/((2x+1+i)(2x+1−i))) =(1/(4(x+((1+i)/2))(x+((1−i)/2)))) =(1/(4(x +(1/(√2)) e^((iπ)/4) )( x +(1/(√2)) e^(−((iπ)/4)) ))) but ((1/(x +(1/(√2))e^(−((iπ)/4)) )) −(1/(x+(1/(√2))e^((iπ)/4) ))) =(1/(√2)) (((2isin((π/4))))/((x+(1/(√2))e^(−((iπ)/4)) )(x+(1/(√2))e^((iπ)/4) ))) ⇒ W(x) =(1/(4i)){ (1/(x+(1/(√2))e^(−((iπ)/4)) )) −(1/(x+(1/(√2))e^((iπ)/4) ))} ⇒ W^((n−1)) (x)=(1/(4i)){ (((−1)^(n−1) (n−1)!)/((x+(1/(√2))e^(−((iπ)/4)) )^n )) −(((−1)^(n−1) (n−1)!)/((x+(1/(√2))e^((iπ)/4) )^n ))} ⇒ f^((n)) (x)=(((−1)^(n−1) (n−1)!)/(2i)){ (1/((x+(1/(√2))e^(−((iπ)/4)) )^n )) −(1/((x+(1/(√2))e^((iπ)/4) )^n ))} and f^((n)) (0) =(((−1)^(n−1) (n−1)!)/(2i)){ (1/((√2))^(−n) e^(−i((nπ)/4)) )) −(1/(((√2))^(−n) e^(i((nπ)/4)) ))} =(((−1)^(n−1) (n−1)!)/(2i)){ ((√2))^n e^((inπ)/4) − ((√2))^n e^(−((inπ)/4)) } =(((−1)^(n−1) (n−1)!)/(2i)) ((√2))^n 2i sin(((nπ)/4)) =(−1)^(n−1) (n−1)! ((√2))^n sin(((nπ)/4))

$${let}\:{f}\left({x}\right)\:={arctan}\left(\mathrm{1}+\mathrm{2}{x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=\frac{\mathrm{2}}{\mathrm{1}+\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{2}} }\:\Rightarrow\:{f}^{\left({n}\right)} \left({x}\right)\:=\mathrm{2}\:\left\{\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right\}^{\left({n}−\mathrm{1}\right)} \:\:{with}\:{n}>\mathrm{0} \\ $$$${let}\:{W}\left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow{W}\left({x}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}+{i}\right)\left(\mathrm{2}{x}+\mathrm{1}−{i}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\left({x}+\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}\left({x}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left(\:{x}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)}\:{but} \\ $$$$\left(\frac{\mathrm{1}}{{x}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\frac{\left(\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right)}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)}\:\Rightarrow \\ $$$${W}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{{x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\ $$$${W}^{\left({n}−\mathrm{1}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\:\:\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\:−\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{{n}} }\right\}\:{and} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\:\frac{\mathrm{1}}{\left.\sqrt{\mathrm{2}}\right)^{−{n}} \:{e}^{−{i}\frac{{n}\pi}{\mathrm{4}}} }\:−\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\right)^{−{n}} \:{e}^{{i}\frac{{n}\pi}{\mathrm{4}}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{\frac{{in}\pi}{\mathrm{4}}} \:−\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{−\frac{{in}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:\mathrm{2}{i}\:{sin}\left(\frac{{n}\pi}{\mathrm{4}}\right)\:=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:\left(\sqrt{\mathrm{2}}\right)^{{n}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$

Question Number 53950    Answers: 0   Comments: 1

calculate ∫_(1/3) ^(1/2) Γ(x)Γ(1−x)dx with Γ(x) =∫_0 ^∞ t^(x−1) e^(−t) dt with x>0 .

$$\:{calculate}\:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right){dx}\:\:\:{with}\:\Gamma\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} {dt}\:\:\:\:{with}\:{x}>\mathrm{0}\:. \\ $$

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