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IntegrationQuestion and Answers: Page 114

Question Number 128529 Answers: 1 Comments: 2

Question Number 128511 Answers: 1 Comments: 0

H=∫ ((2017x^(2016) +2018x^(2017) )/(1+x^(4034) +2x^(4035) +x^(4036) )) dx

$$\:\mathcal{H}=\int\:\frac{\mathrm{2017x}^{\mathrm{2016}} +\mathrm{2018x}^{\mathrm{2017}} }{\mathrm{1}+\mathrm{x}^{\mathrm{4034}} +\mathrm{2x}^{\mathrm{4035}} +\mathrm{x}^{\mathrm{4036}} }\:\mathrm{dx}\: \\ $$

Question Number 128499 Answers: 1 Comments: 0

find u_n =∫_1 ^∞ (([ne^(−x) ])/n^3 )dx

$$\mathrm{find}\:\:\mathrm{u}_{\mathrm{n}} =\int_{\mathrm{1}} ^{\infty} \:\:\frac{\left[\mathrm{ne}^{−\mathrm{x}} \right]}{\mathrm{n}^{\mathrm{3}} }\mathrm{dx} \\ $$

Question Number 128495 Answers: 0 Comments: 0

find ∫_0 ^∞ ((sinx)/([x]))dx

$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sinx}}{\left[\mathrm{x}\right]}\mathrm{dx} \\ $$

Question Number 128471 Answers: 2 Comments: 0

... calculus ... Φ=^? ∫_0 ^( 1) (ln(x))^2 ln((√(−ln(x))) dx

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\:{calculus}\:... \\ $$$$\:\:\:\:\:\:\:\Phi\overset{?} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({ln}\left({x}\right)\right)^{\mathrm{2}} {ln}\left(\sqrt{−{ln}\left({x}\right)}\:{dx}\right. \\ $$

Question Number 128417 Answers: 1 Comments: 0

η = ∫_0 ^( 1) x^3 (1−x^3 )^(n−1) dx

$$\:\:\:\:\:\:\:\:\:\eta\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{x}^{\mathrm{3}} \:\left(\mathrm{1}−\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{n}−\mathrm{1}} \:\mathrm{dx}\: \\ $$

Question Number 128408 Answers: 1 Comments: 0

ρ = ∫ ((sin (4x))/(sin^4 (x)+cos^4 (x))) dx

$$\rho\:=\:\int\:\frac{\mathrm{sin}\:\left(\mathrm{4}{x}\right)}{\mathrm{sin}\:^{\mathrm{4}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{4}} \left({x}\right)}\:{dx}\: \\ $$

Question Number 128385 Answers: 1 Comments: 0

∫_1 ^( π) determinant ((x^3 ,(lnx),(sinx)),((3x^2 ),(1/x),(cosx)),(6,(2x^(−3) ),(−cosx)))dx =?

$$\int_{\mathrm{1}} ^{\:\pi} \begin{vmatrix}{\mathrm{x}^{\mathrm{3}} }&{\mathrm{lnx}}&{\mathrm{sinx}}\\{\mathrm{3x}^{\mathrm{2}} }&{\frac{\mathrm{1}}{\mathrm{x}}}&{\mathrm{cosx}}\\{\mathrm{6}}&{\mathrm{2x}^{−\mathrm{3}} }&{−\mathrm{cosx}}\end{vmatrix}\mathrm{dx}\:=?\: \\ $$

Question Number 128373 Answers: 1 Comments: 0

∫_0 ^1 x^(3/2) (1−x)^(1/2) dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:{dx} \\ $$

Question Number 128334 Answers: 1 Comments: 0

Ω = ∫ (x^2 /( (√((a+bx^2 )^5 )))) dx ; where : a; b >0

$$\Omega\:=\:\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{a}+\mathrm{bx}^{\mathrm{2}} \right)^{\mathrm{5}} }}\:\mathrm{dx}\:;\:\mathrm{where}\::\:\mathrm{a};\:\mathrm{b}\:>\mathrm{0}\: \\ $$

Question Number 128346 Answers: 2 Comments: 0

Question Number 128316 Answers: 1 Comments: 0

nice calculus Ω= ∫_0 ^( ∞) ((sin^3 (x))/x^2 )dx=?

$$\:\:\:\:\:\:\:\:\:\:\:\:{nice}\:\:{calculus} \\ $$$$\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{3}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}=? \\ $$$$ \\ $$

Question Number 128285 Answers: 1 Comments: 0

cos ((π/7))−cos (((2π)/7))+cos (((3π)/7)) =?

$$\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{7}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{7}}\right)\:=? \\ $$

Question Number 128276 Answers: 1 Comments: 0

∫_e^2 ^( ∞) (dx/(x^3 ln x)) ?

$$\:\int_{{e}^{\mathrm{2}} } ^{\:\infty} \:\frac{{dx}}{{x}^{\mathrm{3}} \:\mathrm{ln}\:{x}}\:? \\ $$

Question Number 128262 Answers: 2 Comments: 0

Question Number 128251 Answers: 1 Comments: 0

Question Number 128244 Answers: 0 Comments: 0

...nice calculus... prove that ::Ω= ∫_0 ^(π/4) ln(sin(x))d=((−π)/4)log(2)−(G/2) log(2sin(x))=Σ_(n=1) ^∞ ((−1)/n)cos(2nx) Ω= ∫_0 ^( (π/4)) {−log(2)−Σ_(n=1) ^∞ ((cos(2nx))/n)}dx =((−π)/4)log(2)−Σ_(n=1) ^∞ ∫_0 ^( (π/4)) ((cos(2nx))/n)dx =((−π)/4)log(2)−Σ_(n=1) ^∞ [(1/(2n^2 ))sin(2nx)]_0 ^(π/4) =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ ((sin(((nπ)/2)))/n^2 ) =((−π)/4)log(2)−(1/2){(1/1^2 )−(1/3^2 ) +(1/5^2 )−..} =((−π)/4)log(2)−(1/2)Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 )) ∴ Ω =((−π)/4)log(2)−(G/2) ✓ G:= catalan constant...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:\:\:{prove}\:\:{that}\:\:::\Omega=\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({sin}\left({x}\right)\right){d}=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}} \\ $$$$\:\:\:\:{log}\left(\mathrm{2}{sin}\left({x}\right)\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{1}}{{n}}{cos}\left(\mathrm{2}{nx}\right) \\ $$$$\:\Omega=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \left\{−{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\right\}{dx} \\ $$$$=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}{dx} \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }{sin}\left(\mathrm{2}{nx}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\:\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} } \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }−..\right\} \\ $$$$\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\Omega\:=\frac{−\pi}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{{G}}{\mathrm{2}}\:\:\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{G}:=\:{catalan}\:\:{constant}... \\ $$

Question Number 128221 Answers: 1 Comments: 0

Given f(x+(1/x)) = x^4 −(1/x^4 )+2 then ∫_1 ^( 2) (1−x^(−2) )f(x)dx=

$$\:{Given}\:{f}\left({x}+\frac{\mathrm{1}}{{x}}\right)\:=\:{x}^{\mathrm{4}} −\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\mathrm{2} \\ $$$$\:{then}\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{−\mathrm{2}} \right){f}\left({x}\right){dx}= \\ $$

Question Number 128194 Answers: 1 Comments: 2