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Question Number 95951 by me2love2math last updated on 28/May/20

Commented by me2love2math last updated on 28/May/20

pls help out on this

$${pls}\:{help}\:{out}\:{on}\:{this}\: \\ $$

Commented by prakash jain last updated on 28/May/20

∫(dx/(x^2 (√(4−x^2 )))) x=2sin θ dx=2cos θdθ ∫((2cos θdθ)/((2sin θ)^2 (√(4−4sin^2 θ )))) =∫((2cos θdθ)/(4sin^2 θ∙2cos θ)) =(1/4)∫cosec^2 θdθ =−(1/4)cot θ+C =−(1/4)×((cos θ)/(sin θ))+C =−(1/4)×((√(1−sin^2 θ))/(sin θ))+C =−(1/4)×((√(1−((x/2))^2 ))/(x/2))+C =−(1/4)×((√(4−x^2 ))/x)+C Note: cos θ=(√(1−sin^2 θ)) when −(π/2)<θ<(π/2)

$$\int\frac{{dx}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \\ $$$${x}=\mathrm{2sin}\:\theta \\ $$$${dx}=\mathrm{2cos}\:\theta{d}\theta \\ $$$$\int\frac{\mathrm{2cos}\:\theta{d}\theta}{\left(\mathrm{2sin}\:\theta\right)^{\mathrm{2}} \sqrt{\mathrm{4}−\mathrm{4sin}^{\mathrm{2}} \theta\:}} \\ $$$$=\int\frac{\mathrm{2cos}\:\theta{d}\theta}{\mathrm{4sin}^{\mathrm{2}} \theta\centerdot\mathrm{2cos}\:\theta}\:\:\:\:\:\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{cosec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cot}\:\theta+{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}+{C} \\ $$$$\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}}{\mathrm{sin}\:\theta}+{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\frac{{x}}{\mathrm{2}}}+{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}×\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{{x}}+{C} \\ $$$$ \\ $$$${Note}:\:\mathrm{cos}\:\theta=\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}\:{when}\:−\frac{\pi}{\mathrm{2}}<\theta<\frac{\pi}{\mathrm{2}} \\ $$

Commented by me2love2math last updated on 28/May/20

Thanks...pls Q2 and Q3 in particular

$${Thanks}...{pls}\:{Q}\mathrm{2}\:{and}\:{Q}\mathrm{3}\:{in}\:{particular} \\ $$

Answered by abdomathmax last updated on 28/May/20

I =∫ (dx/(x^2 (√(4−x^2 )))) changement x =2sinθ give I =∫ ((2cosθ dθ)/(4sin^2 θ(2 cosθ))) =∫ (dθ/(4×((1−cos(2θ))/2))) =∫ (dθ/(2−2cos(2θ))) =_(tan(θ)=t) ∫ (dt/((1+t^2 )(2−2×((1−t^2 )/(1+t^2 ))))) =∫ (dt/(2+2t^2 −2+2t^2 )) =∫ (dt/(4t^2 )) =−(1/(4t)) +c =−(1/(4tanθ))+c =−((cosθ)/(4sinθ))+c =−((√(1−sin^2 θ))/(2x))+c =−((√(1−(x^2 /4)))/(2x)) +c =−((√(4−x^2 ))/(4x)) +c

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}\:\mathrm{changement}\:\mathrm{x}\:=\mathrm{2sin}\theta\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int\:\:\frac{\mathrm{2cos}\theta\:\mathrm{d}\theta}{\mathrm{4sin}^{\mathrm{2}} \theta\left(\mathrm{2}\:\mathrm{cos}\theta\right)}\:=\int\:\:\frac{\mathrm{d}\theta}{\mathrm{4}×\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}} \\ $$$$=\int\:\:\frac{\mathrm{d}\theta}{\mathrm{2}−\mathrm{2cos}\left(\mathrm{2}\theta\right)}\:=_{\mathrm{tan}\left(\theta\right)=\mathrm{t}} \:\:\:\:\int\:\:\:\frac{\mathrm{dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{2}−\mathrm{2}×\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\:\frac{\mathrm{dt}}{\mathrm{2}+\mathrm{2t}^{\mathrm{2}} −\mathrm{2}+\mathrm{2t}^{\mathrm{2}} }\:=\int\:\frac{\mathrm{dt}}{\mathrm{4t}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{4t}}\:+\mathrm{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4tan}\theta}+\mathrm{c}\:=−\frac{\mathrm{cos}\theta}{\mathrm{4sin}\theta}+\mathrm{c}\:=−\frac{\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}}{\mathrm{2x}}+\mathrm{c} \\ $$$$=−\frac{\sqrt{\mathrm{1}−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}}}{\mathrm{2x}}\:+\mathrm{c}\:=−\frac{\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{4x}}\:+\mathrm{c} \\ $$

Commented by me2love2math last updated on 28/May/20

thx...Q2 and Q3 is where i have issues

$${thx}...{Q}\mathrm{2}\:{and}\:{Q}\mathrm{3}\:{is}\:{where}\:{i}\:{have}\:{issues} \\ $$

Answered by Sourav mridha last updated on 28/May/20

∫(dx/(x^2 (√(4−x^2 )))) without any substitution−−− ∫((1/x^3 )/(√((4/x^2 )−1)))dx=(1/((−8)))∫((d((4/x^2 )−1))/(√((4/x^2 )−1))) =−(1/4)(√((4/x^2 )−1)) +c

$$\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{without}\:\mathrm{any}\:\mathrm{substitution}−−−\:\:\: \\ $$$$\:\:\:\:\:\int\frac{\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }}{\sqrt{\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}}}\mathrm{dx}=\frac{\mathrm{1}}{\left(−\mathrm{8}\right)}\int\frac{\mathrm{d}\left(\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\right)}{\sqrt{\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}}} \\ $$$$\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}}\:\:\:\:+\mathrm{c} \\ $$

Answered by john santu last updated on 29/May/20

(2) ∫_0 ^1 [∫_x ^(√x) (60x^2 y−40y^3 )dy] dx the inner bracketed integration is carried out first and evaluated as [ 30x^2 y^2 −10y^4 ]_x ^(√x) = 30x^2 (x−x^2 )−10(x^2 −x^4 ) = 30x^3 −30x^4 −10x^2 +10x^4 = 30x^3 −20x^4 −10x^2 so the integration becomes ∫_0 ^1 (30x^3 −20x^4 −10x^2 ) dx = [((30)/4)x^4 −4x^5 −((10)/3)x^3 ]_0 ^1 = ((15)/2)−4−((10)/3) = ((45−24−20)/6) = (1/6) . done

$$\left(\mathrm{2}\right)\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left[\underset{\mathrm{x}} {\overset{\sqrt{\mathrm{x}}} {\int}}\:\left(\mathrm{60x}^{\mathrm{2}} \mathrm{y}−\mathrm{40y}^{\mathrm{3}} \right)\mathrm{dy}\right]\:\mathrm{dx}\: \\ $$$$\mathrm{the}\:\mathrm{inner}\:\mathrm{bracketed}\:\mathrm{integration} \\ $$$$\mathrm{is}\:\mathrm{carried}\:\mathrm{out}\:\mathrm{first}\:\mathrm{and}\:\mathrm{evaluated} \\ $$$$\mathrm{as}\:\left[\:\mathrm{30x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} −\mathrm{10y}^{\mathrm{4}} \:\right]_{\mathrm{x}} ^{\sqrt{\mathrm{x}}} =\: \\ $$$$\mathrm{30x}^{\mathrm{2}} \left(\mathrm{x}−\mathrm{x}^{\mathrm{2}} \right)−\mathrm{10}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{4}} \right)\:= \\ $$$$\mathrm{30x}^{\mathrm{3}} −\mathrm{30x}^{\mathrm{4}} −\mathrm{10x}^{\mathrm{2}} +\mathrm{10x}^{\mathrm{4}} = \\ $$$$\mathrm{30x}^{\mathrm{3}} −\mathrm{20x}^{\mathrm{4}} −\mathrm{10x}^{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{becomes}\: \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{30x}^{\mathrm{3}} −\mathrm{20x}^{\mathrm{4}} −\mathrm{10x}^{\mathrm{2}} \right)\:\mathrm{dx}\:= \\ $$$$\left[\frac{\mathrm{30}}{\mathrm{4}}\mathrm{x}^{\mathrm{4}} −\mathrm{4x}^{\mathrm{5}} −\frac{\mathrm{10}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} = \\ $$$$\frac{\mathrm{15}}{\mathrm{2}}−\mathrm{4}−\frac{\mathrm{10}}{\mathrm{3}}\:=\:\frac{\mathrm{45}−\mathrm{24}−\mathrm{20}}{\mathrm{6}}\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{6}}\:.\:\mathrm{done}\:\: \\ $$

Commented by me2love2math last updated on 29/May/20

thx so much

$${thx}\:{so}\:{much} \\ $$

Answered by john santu last updated on 29/May/20

(3)f(x,y) = 2xy−(1/2)(x^2 +y^2 )+x^2 f(x,y) = 2xy +(1/2)x^2 −(1/2)y^2 firstly find all first and second partial derivatives of f(x,y) . they are f_x =2y+x ; f_y =2x−y f_(xx) = 1 ; f_(yy) = −1 f_(xy) =f_(yx) = 2 setting f_x =0 , yields x=−2y substitution into f_y yields −5y=0 y=0 . this signal possible extrema at (0,0) as x=−2y evaluate D at (0,0) D = 4y^2 −4((1/2))(−(1/2)y^2 ) D = 0 , there is an extremum and as f_(xx) > 0 it is a minimum therefore f(0,0) = 0 is a local minimum. (done)

$$\left(\mathrm{3}\right)\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{2xy}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)+\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{2xy}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{firstly}\:\mathrm{find}\:\mathrm{all}\:\mathrm{first}\:\mathrm{and}\:\mathrm{second}\: \\ $$$$\mathrm{partial}\:\mathrm{derivatives}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:.\:\mathrm{they} \\ $$$$\mathrm{are}\:{f}_{{x}} =\mathrm{2}{y}+{x}\:;\:{f}_{{y}} =\mathrm{2}{x}−{y} \\ $$$${f}_{{xx}} \:=\:\mathrm{1}\:;\:{f}_{{yy}} \:=\:−\mathrm{1} \\ $$$${f}_{{xy}} ={f}_{{yx}} \:=\:\mathrm{2} \\ $$$${setting}\:{f}_{{x}} =\mathrm{0}\:,\:\mathrm{yields}\:{x}=−\mathrm{2}{y} \\ $$$${substitution}\:{into}\:{f}_{{y}} \:{yields}\:−\mathrm{5}{y}=\mathrm{0} \\ $$$${y}=\mathrm{0}\:.\:{this}\:{signal}\:{possible}\:{extrema} \\ $$$${at}\:\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{as}\:\mathrm{x}=−\mathrm{2y} \\ $$$$\mathrm{evaluate}\:\mathrm{D}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right)\: \\ $$$$\mathrm{D}\:=\:\mathrm{4y}^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\mathrm{D}\:=\:\mathrm{0}\:,\:\mathrm{there}\:\mathrm{is}\:\mathrm{an}\:\mathrm{extremum} \\ $$$$\mathrm{and}\:\mathrm{as}\:{f}_{{xx}} \:>\:\mathrm{0}\:{it}\:{is}\:{a}\:{minimum} \\ $$$${therefore}\:{f}\left(\mathrm{0},\mathrm{0}\right)\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{local} \\ $$$$\mathrm{minimum}.\:\left(\mathrm{done}\right) \\ $$

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