Calculate-9-x-2-x-6-dx-

Question Number 76892 by john santu last updated on 31/Dec/19

$$\mathcal{C}{alculate}\:\int\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{6}} }\:{dx}\:. \\ $$

Answered by jagoll last updated on 31/Dec/19

Answered by MJS last updated on 01/Jan/20

∫((√(9−x^2 ))/x^6 )dx= [t=(x/( (√(9−x^2 )))) → dx=(((9−x^2 )^(3/2) )/9)dt] =(1/(81))∫(dt/t^4 )+(1/(81))∫(dt/t^6 )=−(1/(243t^3 ))−(1/(405t^5 ))= =−(((9−x^2 )^(3/2) )/(243x^3 ))−(((9−x^2 )^(5/2) )/(405x^5 ))=−(((2x^2 +27)(9−x^2 )^(3/2) )/(1215x^5 ))+C

$$\int\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{6}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}\:\rightarrow\:{dx}=\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{9}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{81}}\int\frac{{dt}}{{t}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{81}}\int\frac{{dt}}{{t}^{\mathrm{6}} }=−\frac{\mathrm{1}}{\mathrm{243}{t}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{405}{t}^{\mathrm{5}} }= \\ $$$$=−\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{243}{x}^{\mathrm{3}} }−\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{405}{x}^{\mathrm{5}} }=−\frac{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{27}\right)\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{1215}{x}^{\mathrm{5}} }+{C} \\ $$

Commented by jagoll last updated on 01/Jan/20

sorry sir. the equation ∫ ((√(9−x^2 ))/x^6 ) dx .

$${sorry}\:{sir}.\:{the}\:{equation}\:\int\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{6}} }\:{dx}\:. \\ $$

Commented by MJS last updated on 01/Jan/20

thank you, it′s just a typo, I corrected it

$$\mathrm{thank}\:\mathrm{you},\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\ $$

Commented by jagoll last updated on 01/Jan/20

$${thanks}\:{you}\:{sir} \\ $$

Answered by petrochengula last updated on 03/Jan/20

let x=3sinθ⇒dx=3cosθdθ =∫((√(9−9sin^2 θ))/(3^6 sin^6 θ))3cosθdθ=(1/3^4 )∫((cos^2 θ)/(sin^6 θ))dθ =(1/(81))∫cot^2 θcosec^4 θdθ =(1/(81))∫cot^2 θ(1+cot^2 θ)cosec^2 θdθ let t=cotθ⇒−dt=cosec^2 θdθ =−(1/(81))∫t^2 (1+t^2 )dt =−(t^3 /(243))−(t^5 /(405))+c =−((cot^3 θ)/(243))−((cot^5 θ)/(405))+c =−(((cot(sin^(−1) ((x/3))))^3 )/(243))−(((cot(sin^(−1) ((x/3))))^5 )/(405))+c

$${let}\:{x}=\mathrm{3}{sin}\theta\Rightarrow{dx}=\mathrm{3}{cos}\theta{d}\theta \\ $$$$=\int\frac{\sqrt{\mathrm{9}−\mathrm{9}{sin}^{\mathrm{2}} \theta}}{\mathrm{3}^{\mathrm{6}} {sin}^{\mathrm{6}} \theta}\mathrm{3}{cos}\theta{d}\theta=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\int\frac{{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{6}} \theta}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{81}}\int{cot}^{\mathrm{2}} \theta{cosec}^{\mathrm{4}} \theta{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{81}}\int{cot}^{\mathrm{2}} \theta\left(\mathrm{1}+{cot}^{\mathrm{2}} \theta\right){cosec}^{\mathrm{2}} \theta{d}\theta \\ $$$${let}\:{t}={cot}\theta\Rightarrow−{dt}={cosec}^{\mathrm{2}} \theta{d}\theta \\ $$$$=−\frac{\mathrm{1}}{\mathrm{81}}\int{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt} \\ $$$$=−\frac{{t}^{\mathrm{3}} }{\mathrm{243}}−\frac{{t}^{\mathrm{5}} }{\mathrm{405}}+{c} \\ $$$$=−\frac{{cot}^{\mathrm{3}} \theta}{\mathrm{243}}−\frac{{cot}^{\mathrm{5}} \theta}{\mathrm{405}}+{c} \\ $$$$=−\frac{\left({cot}\left({sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{3}}\right)\right)\right)^{\mathrm{3}} }{\mathrm{243}}−\frac{\left({cot}\left({sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{3}}\right)\right)\right)^{\mathrm{5}} }{\mathrm{405}}+{c} \\ $$

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