Question and Answers Forum

All Questions   Topic List

IntegrationQuestion and Answers: Page 104

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

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

Question Number 133963    Answers: 1   Comments: 0

Y = ∫ (dx/( ((1+x^6 ))^(1/6) ))?

$$\mathcal{Y}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{6}}]{\mathrm{1}+{x}^{\mathrm{6}} }}? \\ $$

Question Number 133957    Answers: 2   Comments: 0

1)decompose F(x)=(1/((x+1)^5 (2x−1)^4 )) 2) find ∫_1 ^∞ F(x)dx

$$\left.\mathrm{1}\right)\mathrm{decompose}\:\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\int_{\mathrm{1}} ^{\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx} \\ $$

Question Number 133943    Answers: 0   Comments: 0

.......advanced calculus...... prove that: 𝛗= ∫_0 ^( ∞) ((cos(x^2 )−cos(x))/x)dx=(γ/2) γ: euler−mascheroni constant...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{advanced}\:\:\:\:\:{calculus}...... \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({x}^{\mathrm{2}} \right)−{cos}\left({x}\right)}{{x}}{dx}=\frac{\gamma}{\mathrm{2}} \\ $$$$\:\:\:\gamma:\:{euler}−{mascheroni}\:{constant}... \\ $$

Question Number 133911    Answers: 1   Comments: 0

A=∫ ((cos x)/(sin 5x+sin x)) dx ?

$$\mathcal{A}=\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{5x}+\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Question Number 133910    Answers: 0   Comments: 0

De^ montrer que; Σ_(n=1) ^∞ (((−1)^n )/(1+n^4 ))=(1/2)[(π/( (√2))) ((sin((π/( (√2))))cosh((π/( (√2))))+sinh((π/( (√2))))cos((π/( (√2)))))/(sinh^2 ((π/( (√2))))+sin^2 ((π/( (√2))))))]

$$\:\:\:\:\mathcal{D}\acute {\mathrm{e}montrer}\:\mathrm{que}; \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{1}+\mathrm{n}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cosh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sinh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{sinh}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}\right] \\ $$

Question Number 133907    Answers: 0   Comments: 0

Question Number 133938    Answers: 1   Comments: 0

Question Number 133825    Answers: 1   Comments: 0

Question Number 133791    Answers: 2   Comments: 0

......advanced integral.... prove that :: 𝛗=∫_0 ^( ∞) (((1−e^(−ϕx) )/(1+e^(ϕx) )) )(dx/x) =?? ϕ: = Golden ratio...

$$\:\:\:\:\:\:\:\:\:\:\:\:......{advanced}\:\:\:\:{integral}.... \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}−{e}^{−\varphi{x}} }{\mathrm{1}+{e}^{\varphi{x}} }\:\right)\frac{{dx}}{{x}}\:=?? \\ $$$$\:\:\:\varphi:\:=\:{Golden}\:{ratio}... \\ $$$$ \\ $$

Question Number 133786    Answers: 1   Comments: 0

A = ∫_0 ^( 1) sin^(−1) (((x^2 +1)/( (√(2x^4 +2)))) ) dx =?

$$\mathcal{A}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{sin}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\:\sqrt{\mathrm{2}{x}^{\mathrm{4}} +\mathrm{2}}}\:\right)\:{dx}\:=? \\ $$

Question Number 133951    Answers: 3   Comments: 0

V = ∫ ln (x+(√(1+x^2 )) ) dx

$$\:\mathcal{V}\:=\:\int\:\mathrm{ln}\:\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right)\:\mathrm{dx}\: \\ $$

Question Number 133719    Answers: 1   Comments: 1

Question Number 133725    Answers: 0   Comments: 7

  Pg 99      Pg 100      Pg 101      Pg 102      Pg 103      Pg 104      Pg 105      Pg 106      Pg 107      Pg 108   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com