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IntegrationQuestion and Answers: Page 202

Question Number 81427 Answers: 1 Comments: 4

1) calculate ∫_0 ^∞ cos(x^3 )dx 2)find the value of ∫_0 ^∞ cos(x^n )dx with n≥2

$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{\mathrm{3}} \right){dx} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}^{{n}} \right){dx}\:{with}\:{n}\geqslant\mathrm{2} \\ $$

Question Number 81400 Answers: 1 Comments: 3

Hello Nice day im thinking of this one a close forme? ∫(√(1+x^p ))dx p∈R_+ , x∈[0,1[

$${Hello}\:\:{Nice}\:{day}\:{im}\:{thinking}\:{of}\:{this}\:{one}\:\:{a}\:{close}\:{forme}? \\ $$$$\int\sqrt{\mathrm{1}+{x}^{{p}} }{dx} \\ $$$${p}\in\mathbb{R}_{+} , \\ $$$${x}\in\left[\mathrm{0},\mathrm{1}\left[\right.\right. \\ $$

Question Number 81340 Answers: 1 Comments: 4

Quation posted Times ago i can′t find it .after Somme Try i got close Fofme ∫_0 ^π sin(x^2 )dx=(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4),−(π^4 /4))? sin(x)=Σ_(k≥0) (((−1)^k x^(2k+1) )/((2k+1)!)) ⇒sin(x^2 )=Σ_(k≥0) (((−1)^k x^(4k+2) )/((2k+1)!)) ∫_0 ^π sin(x^2 )dx=Σ_(k≥0) (((−1)^k )/((2k+1)!))∫_0 ^π x^(4k+2) dx =Σ_(k≥0) (((−1)^k π^(4k+3) )/((2k+1)!(4k+3))) =(π^3 /3)Σ_(k≥0) ((3(−1)^k π^(4k) )/((2k+1)!(4k+3))) (2k+1)!=k!.Π_(j=1) ^(k+1) (k+j)=k!.2^k .3....(2k+1) =(π^3 /3).Σ_(k≥0) ((3(−1)^k π^(4k) )/(k!.2^k .3....(2k+1)(4k+3))) =(π^3 /3)Σ_(k≥0) ((2^k .3)/(3...(2k+1).(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) ((2^k .3.....(4k−1))/(3.....(2k+1).(7.....(4k−1)(4k+3))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3).Σ_(k≥0) (((1/4^k ).(3)...(4k−1))/((1/2^k ).(3)...(2k+1).(1/4^k )(7)...(4k+3))).(−(π^4 /4))^k .(1/(k!)) =(π^3 /3)Σ_(k≥0) ((((3/4))....((3/4)+k−1))/(((3/2))....((3/2)+k−1).((7/4))....((7/4)+k−1))).(((−π^4 )/4))^k .(1/(k!)) =(π^3 /3) _1 F_2 ((3/4);(3/2);(7/4);−(π^4 /4))

Question Number 81336 Answers: 1 Comments: 7

decompose F(x)=(1/((x−1)^3 (x+3)^7 )) and detrmine ∫ F(x)dx 2) calculate ∫_2 ^(+∞) F(x)dx

$${decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{3}\right)^{\mathrm{7}} } \\ $$$${and}\:{detrmine}\:\:\int\:{F}\left({x}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{2}} ^{+\infty} \:{F}\left({x}\right){dx} \\ $$

Question Number 81313 Answers: 1 Comments: 5

∫(((x−2)^3 )/((x+2)^5 )) dx

$$\int\frac{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} }{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{5}} }\:\mathrm{dx} \\ $$

Question Number 81290 Answers: 0 Comments: 3

∫_(−1) ^1 arctan(x)arctan((x/(√(1−x^2 ))))arctan(((1+x)/(1−x)))dx

$$\int_{−\mathrm{1}} ^{\mathrm{1}} {arctan}\left({x}\right){arctan}\left(\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right){arctan}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right){dx} \\ $$

Question Number 81279 Answers: 0 Comments: 2

∫ ((√(2x+1))/(3x)) dx = ?

$$\int\:\frac{\sqrt{\mathrm{2}{x}+\mathrm{1}}}{\mathrm{3}{x}}\:{dx}\:=\:? \\ $$

Question Number 81165 Answers: 1 Comments: 3

let f(x)=(1/(x^2 −2(cosθ)x +1)) 1) calculate f^((n)) (x) 2) find ∫_0 ^∞ f(x)dx 3) find ∫_0 ^∞ f^((n)) (x)dx

$${let}\:{f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}\left({cos}\theta\right){x}\:+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\mathrm{0}} ^{\infty} \:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\int_{\mathrm{0}} ^{\infty} \:{f}^{\left({n}\right)} \left({x}\right){dx} \\ $$

Question Number 81100 Answers: 0 Comments: 2

∫ ((x dx)/((tan x+cot x)^2 )) = ?

$$\int\:\frac{{x}\:{dx}}{\left(\mathrm{tan}\:{x}+\mathrm{cot}\:{x}\right)^{\mathrm{2}} }\:=\:? \\ $$

Question Number 81043 Answers: 1 Comments: 8

∫ ((2e^(2x) −e^x )/(√(3e^(2x) −6e^x −1))) dx

$$\int\:\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −{e}^{{x}} }{\sqrt{\mathrm{3}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}}}\:{dx} \\ $$

Question Number 80977 Answers: 1 Comments: 6

Question Number 80951 Answers: 1 Comments: 3

Question Number 80925 Answers: 0 Comments: 1

∫_(−∞) ^∞ ((cos(x))/(1+x^2 )) dx =(π/e)

$$\int_{−\infty} ^{\infty} \frac{{cos}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=\frac{\pi}{{e}} \\ $$

Question Number 80924 Answers: 1 Comments: 3

show that ∫_0 ^∞ (x^((π/5)−1) /(1+x^(2π) )) dx =φ

$${show}\:{that} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\frac{\pi}{\mathrm{5}}−\mathrm{1}} }{\mathrm{1}+{x}^{\mathrm{2}\pi} }\:{dx}\:=\phi\: \\ $$

Question Number 80921 Answers: 0 Comments: 2

∫_0 ^(π/2) ((xdx)/(sin x+cos x)) = ?

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{xdx}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:=\:? \\ $$

Question Number 80914 Answers: 0 Comments: 0

(1) Integrate F(x, y) = x^2 over the region bounded by y = x^2 , x = 2 and x = 1 (2) Integrate G(x, y) = x^2 + y^2 over the region bounded by the triangle x = y, y = 1 and y = 0

$$\left(\mathrm{1}\right) \\ $$$$\mathrm{Integrate}\:\:\mathrm{F}\left(\mathrm{x},\:\mathrm{y}\right)\:\:=\:\:\mathrm{x}^{\mathrm{2}} \:\:\:\mathrm{over}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{x}^{\mathrm{2}} , \\ $$$$\mathrm{x}\:\:=\:\:\mathrm{2}\:\:\mathrm{and}\:\mathrm{x}\:\:=\:\:\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{Integrate}\:\:\:\:\mathrm{G}\left(\mathrm{x},\:\mathrm{y}\right)\:\:=\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:\:\:\:\mathrm{over}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\: \\ $$$$\mathrm{triangle}\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{y},\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{y}\:\:=\:\:\mathrm{0} \\ $$

Question Number 80882 Answers: 1 Comments: 4