# 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
$$\int\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{6}} }{dx}= \\$$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\:\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}\:\rightarrow\:{dx}=\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{9}}{dt}\right] \\$$$$=\frac{\mathrm{1}}{\mathrm{81}}\int\frac{{dt}}{{t}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{81}}\int\frac{{dt}}{{t}^{\mathrm{6}} }=−\frac{\mathrm{1}}{\mathrm{243}{t}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{405}{t}^{\mathrm{5}} }= \\$$$$=−\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{243}{x}^{\mathrm{3}} }−\frac{\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{405}{x}^{\mathrm{5}} }=−\frac{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{27}\right)\left(\mathrm{9}−{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{1215}{x}^{\mathrm{5}} }+{C} \\$$
Commented by jagoll last updated on 01/Jan/20
$${sorry}\:{sir}.\:{the}\:{equation}\:\int\:\frac{\sqrt{\mathrm{9}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{6}} }\:{dx}\:. \\$$
Commented by MJS last updated on 01/Jan/20
$$\mathrm{thank}\:\mathrm{you},\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{I}\:\mathrm{corrected}\:\mathrm{it} \\$$
Commented by jagoll last updated on 01/Jan/20
$${thanks}\:{you}\:{sir} \\$$
Answered by petrochengula last updated on 03/Jan/20
$${let}\:{x}=\mathrm{3}{sin}\theta\Rightarrow{dx}=\mathrm{3}{cos}\theta{d}\theta \\$$$$=\int\frac{\sqrt{\mathrm{9}−\mathrm{9}{sin}^{\mathrm{2}} \theta}}{\mathrm{3}^{\mathrm{6}} {sin}^{\mathrm{6}} \theta}\mathrm{3}{cos}\theta{d}\theta=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\int\frac{{cos}^{\mathrm{2}} \theta}{{sin}^{\mathrm{6}} \theta}{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{81}}\int{cot}^{\mathrm{2}} \theta{cosec}^{\mathrm{4}} \theta{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{81}}\int{cot}^{\mathrm{2}} \theta\left(\mathrm{1}+{cot}^{\mathrm{2}} \theta\right){cosec}^{\mathrm{2}} \theta{d}\theta \\$$$${let}\:{t}={cot}\theta\Rightarrow−{dt}={cosec}^{\mathrm{2}} \theta{d}\theta \\$$$$=−\frac{\mathrm{1}}{\mathrm{81}}\int{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right){dt} \\$$$$=−\frac{{t}^{\mathrm{3}} }{\mathrm{243}}−\frac{{t}^{\mathrm{5}} }{\mathrm{405}}+{c} \\$$$$=−\frac{{cot}^{\mathrm{3}} \theta}{\mathrm{243}}−\frac{{cot}^{\mathrm{5}} \theta}{\mathrm{405}}+{c} \\$$$$=−\frac{\left({cot}\left({sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{3}}\right)\right)\right)^{\mathrm{3}} }{\mathrm{243}}−\frac{\left({cot}\left({sin}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{3}}\right)\right)\right)^{\mathrm{5}} }{\mathrm{405}}+{c} \\$$