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IntegrationQuestion and Answers: Page 104

Question Number 134580 Answers: 2 Comments: 0

∫(√(tan(x)))dx=...?

$$\int\sqrt{{tan}\left({x}\right)}{dx}=...? \\ $$

Question Number 134570 Answers: 0 Comments: 0

∫_0 ^( 1) ∫_y ^( y^(1/3) ) (√(x^4 +1)) dx dy =?

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{y}} ^{\:\mathrm{y}^{\mathrm{1}/\mathrm{3}} } \sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\:\mathrm{dx}\:\mathrm{dy}\:=?\:\: \\ $$

Question Number 134547 Answers: 2 Comments: 0

R = ∫_0 ^( 3a) x (√((3a−x)/(x+a))) dx

$$\mathscr{R}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{3}{a}} {x}\:\sqrt{\frac{\mathrm{3}{a}−{x}}{{x}+{a}}}\:{dx}\: \\ $$

Question Number 134544 Answers: 0 Comments: 0

Question Number 134474 Answers: 3 Comments: 0

Evaluate ∫_0 ^∞ ((sinx)/x^a )dx

$${Evaluate}\:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{{sinx}}{{x}^{{a}} }{dx} \\ $$

Question Number 134456 Answers: 0 Comments: 1

∫^( ∞) _(−∞) ((sin^3 x)/x^3 ) dx ?

$$\underset{−\infty} {\int}^{\:\:\:\:\infty} \frac{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:\mathrm{dx}\:? \\ $$

Question Number 134454 Answers: 0 Comments: 0

...nice calculus... find the value of: 𝛗=Σ_(n=1) ^∞ ((ζ(2n)−1)/(n+1))=?

$$\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:{calculus}... \\ $$$$\:\:\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\zeta\left(\mathrm{2}{n}\right)−\mathrm{1}}{{n}+\mathrm{1}}=? \\ $$$$ \\ $$

Question Number 134420 Answers: 1 Comments: 0

∫^( π/2) _(−π/2) (1/(2019^x +1)). ((sin^(2020) x)/(sin^(2020) x+cos^(2020) x)) dx ?

$$\underset{−\pi/\mathrm{2}} {\int}^{\:\:\:\:\:\:\pi/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2019}^{{x}} +\mathrm{1}}.\:\frac{\mathrm{sin}\:^{\mathrm{2020}} {x}}{\mathrm{sin}\:^{\mathrm{2020}} {x}+\mathrm{cos}\:^{\mathrm{2020}} {x}}\:{dx}\:? \\ $$

Question Number 134418 Answers: 1 Comments: 0

𝛗=∫_0 ^( ∞) (arctan((1/x)))^2 =???

$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)^{\mathrm{2}} =??? \\ $$

Question Number 134303 Answers: 2 Comments: 0

F=∫_0 ^∞ ((16 arctan (x))/(1+x^2 )) dx

$$\mathcal{F}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{16}\:\mathrm{arctan}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$

Question Number 134302 Answers: 2 Comments: 0

∫_0 ^( 1) ((ln x)/( (√(1−x^2 )))) dx

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$

Question Number 134301 Answers: 3 Comments: 0

Ω = ∫_0 ^∞ (x^2 /((1+x^2 )^4 )) dx

$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:{dx} \\ $$

Question Number 134291 Answers: 1 Comments: 0

Prove ∫_0 ^( ∞) (x^a /(1+e^x ))dx = (1−2^(−a) )𝛇(a+1)𝚪(a+1)

$$ \\ $$$$\:\boldsymbol{\mathrm{Prove}}\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\mathrm{dx}\:=\:\left(\mathrm{1}−\mathrm{2}^{−\mathrm{a}} \right)\boldsymbol{\zeta}\left(\mathrm{a}+\mathrm{1}\right)\boldsymbol{\Gamma}\left(\mathrm{a}+\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 134289 Answers: 3 Comments: 0

I=∫ (x^n /( (√(ax+b)))) dx H=∫ (x^4 /( (√(2x+1)))) dx

$$\mathrm{I}=\int\:\frac{\mathrm{x}^{\mathrm{n}} }{\:\sqrt{\mathrm{ax}+\mathrm{b}}}\:\mathrm{dx} \\ $$$$\mathrm{H}=\int\:\frac{\mathrm{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{2x}+\mathrm{1}}}\:\mathrm{dx} \\ $$

Question Number 134272 Answers: 2 Comments: 0

nice calculus prove that: : i :: =∫_0 ^( ∞) ((cos(πx))/(e^(2π(√x) ) −1)) dx=((2−(√2) )/8) ii:: compute: Σ_(n=−∞) ^∞ (1/(n^4 +9n^2 +10)) =? ...m.n...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nice}\:\:\:\mathrm{calculus} \\ $$$$\:\:\:\:\mathrm{prove}\:\mathrm{that}:\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}\:::\:\: =\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\pi{x}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}\:{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}\:}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{compute}:\:\:\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...{m}.{n}... \\ $$

Question Number 134239 Answers: 1 Comments: 2

can i ask for some help? how to prove this? (1/2)<∫_0 ^(1/2) (dx/( (√(1−x^3 ))))<(π/6)

$${can}\:{i}\:{ask}\:{for}\:{some}\:{help}? \\ $$$${how}\:{to}\:{prove}\:{this}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}<\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}<\frac{\pi}{\mathrm{6}} \\ $$

Question Number 134227 Answers: 0 Comments: 1

I struck upon this Σ_(n=0) ^∞ 16^n =((−1)/(15)) ∫_0 ^( π) cos^4 x sin x dx=((−1)/5)[cos^5 x]_0 ^π =0.4 in another way I=∫^ cos^4 x sin x dx I^ =cos^4 (x) (−cos x)−∫4cos^3 x(−sin x)(−cos x)dx I^ =cos^5 x−4∫cos^4 x sin xdx I^ =cos^5 x−4I I=−cos^5 xΣ_(n=0) ^∞ (−4)^n +C [I]_0 ^π =[(−(−1)Σ_(n=0) ^∞ (−4)^n )−(−Σ_(n=0) ^∞ (−4)^n )] =2Σ_(n=0) ^∞ (−4)^n =2(Σ_(n=0) ^∞ 16^n −4Σ_(n=0) ^∞ 16^n ) =2(−3Σ_(n=0) ^∞ 16^n )=−6Σ_(n=0) ^∞ 16^n −6Σ_(n=0) ^∞ 16^n =0.4⇒Σ_(n=0) ^∞ 16^n =((−0.4)/6)=((−1)/(15)) did I do something wrong?

$$\mathrm{I}\:\:\mathrm{struck}\:\mathrm{upon}\:\mathrm{this} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=\frac{−\mathrm{1}}{\mathrm{5}}\left[\mathrm{cos}^{\mathrm{5}} \:{x}\right]_{\mathrm{0}} ^{\pi} =\mathrm{0}.\mathrm{4} \\ $$$$\mathrm{in}\:\mathrm{another}\:\mathrm{way} \\ $$$${I}=\int^{\:} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{4}} \:\left({x}\right)\:\left(−\mathrm{cos}\:{x}\right)−\int\mathrm{4cos}^{\mathrm{3}} \:{x}\left(−\mathrm{sin}\:{x}\right)\left(−\mathrm{cos}\:{x}\right){dx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}−\mathrm{4}\int\mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{xdx} \\ $$$${I}^{\:} =\mathrm{cos}^{\mathrm{5}} \:{x}−\mathrm{4}{I} \\ $$$${I}=−\mathrm{cos}^{\mathrm{5}} \:{x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} +{C} \\ $$$$\left[{I}\right]_{\mathrm{0}} ^{\pi} =\left[\left(−\left(−\mathrm{1}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} \right)−\left(−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} \right)\right] \\ $$$$=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{4}\right)^{{n}} =\mathrm{2}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} −\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right) \\ $$$$=\mathrm{2}\left(−\mathrm{3}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \right)=−\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} \\ $$$$−\mathrm{6}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\mathrm{0}.\mathrm{4}\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{0}.\mathrm{4}}{\mathrm{6}}=\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\mathrm{did}\:\mathrm{I}\:\mathrm{do}\:\mathrm{something}\:\mathrm{wrong}? \\ $$

Question Number 134219 Answers: 1 Comments: 0

∫_0 ^( π) cos^4 x sin x dx=?

$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=? \\ $$

Question Number 134185 Answers: 1 Comments: 0

.....advanced calculus.... prove that:: i:𝛗=∫_0 ^( (π/2)) ((cos(tan(x)−x))/(cos(x)))dx=(π/e) ii:Π_(n=2) ^∞ e(1−(1/n^2 ))^n^2 =(π/(e(√e)))

$$\:\:\:\:\:\:\:\:\:\:\:\:.....{advanced}\:\:\:{calculus}.... \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:{i}:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({tan}\left({x}\right)−{x}\right)}{{cos}\left({x}\right)}{dx}=\frac{\pi}{{e}} \\ $$$$\:\:\:\:{ii}:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}{e}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } =\frac{\pi}{{e}\sqrt{{e}}} \\ $$$$\:\: \\ $$

Question Number 134182 Answers: 1 Comments: 0

Question Number 134058 Answers: 1 Comments: 0

J = ∫ (dx/(1+tan x+csc x+cot x+sec x))

$$\mathcal{J}\:=\:\int\:\frac{{dx}}{\mathrm{1}+\mathrm{tan}\:{x}+\mathrm{csc}\:{x}+\mathrm{cot}\:{x}+\mathrm{sec}\:{x}} \\ $$

Question Number 134039 Answers: 0 Comments: 0

Question Number 134038 Answers: 1 Comments: 0

Question Number 134016 Answers: 1 Comments: 0

?prove :Σ_(n=1) ^∞ (((−1)^(n−1) H_(2n) )/(2n+1))=(π/8)ln(2)..

$$ \\ $$$$\:\:\:\:\:\:?{prove}\::\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {H}_{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}=\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right).. \\ $$

Question Number 133973 Answers: 1 Comments: 0

Given { ((f(x)=((x+(√(x^2 +(1/(27))))))^(1/3) +((x−(√(x^2 +(1/(27))))))^(1/3) )),((g(x)=x^3 +x+1)) :} Find ∫_0 ^4 (g○f○g)(x) dx .

$$\:\mathrm{Given}\:\begin{cases}{\mathrm{f}\left(\mathrm{x}\right)=\sqrt[{\mathrm{3}}]{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{27}}}}}\\{\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{x}+\mathrm{1}}\end{cases} \\ $$$$\mathrm{Find}\:\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:\left(\mathrm{g}\circ\mathrm{f}\circ\mathrm{g}\right)\left(\mathrm{x}\right)\:\mathrm{dx}\:. \\ $$

Question Number 133972 Answers: 1 Comments: 0

H = ∫ (((2x−1)^7 )/((2x+1)^9 )) dx

$$\mathscr{H}\:=\:\int\:\frac{\left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{7}} }{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{9}} }\:\mathrm{dx}\: \\ $$

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