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Question Number 221056    Answers: 1   Comments: 0

Question Number 221052    Answers: 2   Comments: 0

Question Number 221050    Answers: 0   Comments: 1

Question Number 221049    Answers: 0   Comments: 0

∫ ((cos x)/((x + cos x)^2 )) + ((2(1 − sin x)^2 )/((x + cos x)^3 )) dx

$$ \\ $$$$\:\:\:\:\int\:\:\frac{\mathrm{cos}\:{x}}{\left({x}\:+\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:+\:\frac{\mathrm{2}\left(\mathrm{1}\:−\:\mathrm{sin}\:{x}\right)^{\mathrm{2}} }{\left({x}\:+\:\mathrm{cos}\:{x}\right)^{\mathrm{3}} }\:\:\mathrm{d}{x}\: \\ $$$$\: \\ $$

Question Number 221048    Answers: 2   Comments: 0

∫_0 ^(π/2) cosec (x−(π/3))cosec (x−(π/6))dx

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)\mathrm{cosec}\:\left({x}−\frac{\pi}{\mathrm{6}}\right){dx}\: \\ $$

Question Number 221047    Answers: 1   Comments: 0

Question Number 221044    Answers: 0   Comments: 1

If V be a function of x and y, prove that (∂^2 V/∂x^2 )+(∂^2 V/∂y^2 )=(∂^2 V/∂r^2 )+(1/r) (∂V/∂r)+(1/r^2 ) (∂^2 V/∂θ^2 ), where x=r cos θ , y=rsin θ

$${If}\:{V}\:{be}\:{a}\:{function}\:{of}\:{x}\:{and}\:{y},\:{prove}\:{that} \\ $$$$\frac{\partial^{\mathrm{2}} {V}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {V}}{\partial{y}^{\mathrm{2}} }=\frac{\partial^{\mathrm{2}} {V}}{\partial{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}}\:\frac{\partial{V}}{\partial{r}}+\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial^{\mathrm{2}} {V}}{\partial\theta^{\mathrm{2}} }, \\ $$$${where}\:{x}={r}\:\mathrm{cos}\:\theta\:,\:{y}={r}\mathrm{sin}\:\theta \\ $$

Question Number 221036    Answers: 0   Comments: 0

prove ∫_(P∈[ε,ε+δ]) (√((((d )/dt)f(t))^2 +(((d )/dt)g(t))^2 )) dt≤∫_(P∈[ε,ε+δ]) ((C_1 f^((1)) (t)+C_2 g^((1)) (t))/( (√(C_1 ^2 +C_2 ^2 )))) dt

$$\mathrm{prove} \\ $$$$\int_{{P}\in\left[\epsilon,\epsilon+\delta\right]} \sqrt{\left(\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}{f}\left({t}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\mathrm{g}\left({t}\right)\right)^{\mathrm{2}} }\:\mathrm{d}{t}\leq\int_{{P}\in\left[\epsilon,\epsilon+\delta\right]} \:\frac{{C}_{\mathrm{1}} {f}^{\left(\mathrm{1}\right)} \left({t}\right)+{C}_{\mathrm{2}} \mathrm{g}^{\left(\mathrm{1}\right)} \left({t}\right)}{\:\sqrt{{C}_{\mathrm{1}} ^{\mathrm{2}} +{C}_{\mathrm{2}} ^{\mathrm{2}} }}\:\mathrm{d}{t} \\ $$

Question Number 221025    Answers: 1   Comments: 0

Question Number 221017    Answers: 1   Comments: 0

Question Number 221034    Answers: 1   Comments: 0

Question Number 221012    Answers: 3   Comments: 4

Question Number 221008    Answers: 0   Comments: 0

Question Number 220995    Answers: 2   Comments: 0

Question Number 220987    Answers: 1   Comments: 0

Let a,b,c be positive reals such that abc=1.prove that (1/(a^3 (b+c)))+(1/(b^3 (c+a)))+(1/(c^3 (a+b)))≥(3/2)

$${Let}\:{a},{b},{c}\:{be}\:{positive}\:{reals}\:{such}\:{that}\:{abc}=\mathrm{1}.{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} \left({b}+{c}\right)}+\frac{\mathrm{1}}{{b}^{\mathrm{3}} \left({c}+{a}\right)}+\frac{\mathrm{1}}{{c}^{\mathrm{3}} \left({a}+{b}\right)}\geqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Question Number 220973    Answers: 1   Comments: 0

Prove:Σ_(k=0) ^n (−(1/3))^k cos^3 (3^(k−n) π)=(3/4)[(−(1/3))^(n+1) +cos(π/3^n )]

$$\mathrm{Prove}:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{k}} \mathrm{cos}^{\mathrm{3}} \left(\mathrm{3}^{{k}−{n}} \pi\right)=\frac{\mathrm{3}}{\mathrm{4}}\left[\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}+\mathrm{1}} +\mathrm{cos}\frac{\pi}{\mathrm{3}^{{n}} }\right] \\ $$

Question Number 220972    Answers: 1   Comments: 2

∫_(−∞) ^(+∞) ∫_(−∞) ^(+∞) (1/((x^2 +y^2 +4)^2 )) dxdy x=rcos(θ) y=rsin(θ) ∣∣J∣∣=rdrdθ ∫_0 ^( 2π) ∫_0 ^( ∞) (r/((r^2 +4)^2 ))drdθ=(1/8)∫_0 ^( 2π) dθ=(π/4) Q. if ∫_0 ^( ∞) ∫_0 ^( ∞) (1/((x^2 +y^2 +4)^2 )) dxdy 0≤r<∞ , 0≤θ≤(π/2).....??? I really confuse how could i select interval of integral

$$\int_{−\infty} ^{+\infty} \int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$${x}={r}\mathrm{cos}\left(\theta\right) \\ $$$${y}={r}\mathrm{sin}\left(\theta\right) \\ $$$$\mid\mid\boldsymbol{{J}}\mid\mid={r}\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{{r}}{\left({r}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\mathrm{d}{r}\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\mathrm{d}\theta=\frac{\pi}{\mathrm{4}} \\ $$$${Q}.\:\mathrm{if}\:\int_{\mathrm{0}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\mathrm{0}\leq{r}<\infty\:,\:\mathrm{0}\leq\theta\leq\frac{\pi}{\mathrm{2}}.....??? \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{confuse}\:\mathrm{how}\:\mathrm{could}\:\:\mathrm{i}\:\mathrm{select}\:\mathrm{interval}\:\mathrm{of}\:\mathrm{integral} \\ $$

Question Number 220971    Answers: 1   Comments: 6

Question Number 221680    Answers: 2   Comments: 4

I suspect π=i∫_i ^(−1) (((z−1)dz)/( z(√(z^2 +1)))) someone please help confirm or reject!

$${I}\:{suspect} \\ $$$$\pi={i}\underset{\boldsymbol{{i}}} {\overset{−\mathrm{1}} {\int}}\frac{\left({z}−\mathrm{1}\right){dz}}{\:{z}\sqrt{{z}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${someone}\:{please}\:{help}\:{confirm}\:{or}\:{reject}! \\ $$

Question Number 220968    Answers: 0   Comments: 0

Question Number 220964    Answers: 0   Comments: 0

Find the general solution of the differential equation x^2 (d^3 y/dx^3 ) + x(d^2 y/dx^2 )−6(dy/dx)+6(y/x)=((x ln x+1)/x^2 ),[x>0]

$${Find}\:{the}\:{general}\:{solution}\:{of}\:{the}\:{differential}\:{equation} \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }\:+\:{x}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{6}\frac{{dy}}{{dx}}+\mathrm{6}\frac{{y}}{{x}}=\frac{{x}\:\mathrm{ln}\:{x}+\mathrm{1}}{{x}^{\mathrm{2}} },\left[{x}>\mathrm{0}\right] \\ $$

Question Number 220963    Answers: 1   Comments: 0

Let f:R^2 →R be defined by f(x,y)={(y/(sin y_( 1, y=0) )), y≠0 Then the integral (1/π^2 )∫_(x=0) ^1 ∫_(y=sin^(−1) x) ^(π/2) f(x,y)dy dx correct upto three decimal places,is...

$${Let}\:{f}:\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}\:{be}\:{defined}\:{by}\:{f}\left({x},{y}\right)=\left\{\frac{{y}}{\underset{\:\:\mathrm{1},\:{y}=\mathrm{0}} {\mathrm{sin}\:{y}}},\:{y}\neq\mathrm{0}\right. \\ $$$${Then}\:{the}\:{integral}\:\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\underset{{x}=\mathrm{0}} {\overset{\mathrm{1}} {\int}}\underset{{y}=\mathrm{sin}^{−\mathrm{1}} {x}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{f}\left({x},{y}\right){dy}\:{dx}\:{correct}\:{upto}\:{three}\:{decimal}\:{places},{is}... \\ $$

Question Number 220958    Answers: 1   Comments: 1

for x, y, z >0 find the maximum of x^m y^n z^k subject to ax+by+cz=d.

$${for}\:{x},\:{y},\:{z}\:>\mathrm{0}\:{find}\:{the}\:{maximum}\:{of} \\ $$$${x}^{{m}} {y}^{{n}} {z}^{{k}} \:{subject}\:{to}\:{ax}+{by}+{cz}={d}. \\ $$

Question Number 220950    Answers: 1   Comments: 0

∫_( 0) ^( π) ∫_( 0) ^( 1) ∫_( 0) ^( π) sin^( 2) x + y sin z dxdydz = (1/2) π (2 + π)

$$ \\ $$$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\pi} \int_{\:\mathrm{0}} ^{\:\mathrm{1}} \int_{\:\mathrm{0}} ^{\:\:\pi} \:\mathrm{sin}^{\:\mathrm{2}} \:{x}\:+\:{y}\:\mathrm{sin}\:{z}\:{dxdydz}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\pi\:\left(\mathrm{2}\:+\:\pi\right)\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220948    Answers: 1   Comments: 0

∫ x^2 (√(5−x^6 ))dx

$$\int\:{x}^{\mathrm{2}} \sqrt{\mathrm{5}−{x}^{\mathrm{6}} }{dx} \\ $$

Question Number 220947    Answers: 1   Comments: 0

Σ_(k=1) ^(13) (1/(sin ((π/4)+(((k−1)π)/6))sin ((π/4)+((kπ)/6))))

$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)} \\ $$

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