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

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

Answered by jagoll last updated on 31/Dec/19

Answered by MJS last updated on 01/Jan/20

$$\int \frac{\sqrt{9-x^2}}{x^6}dx=$$$$[t=\frac{x}{\sqrt{9-x^2}}\to dx=\frac{{(9-x^2)}^{\frac{3}{2}}}{9}dt]$$$$=\frac{1}{81}\int \frac{dt}{t^4}+\frac{1}{81}\int \frac{dt}{t^6}=-\frac{1}{243t^3}-\frac{1}{405t^5}=$$$$=-\frac{{(9-x^2)}^{\frac{3}{2}}}{243x^3}-\frac{{(9-x^2)}^{\frac{5}{2}}}{405x^5}=-\frac{(2x^2+27){(9-x^2)}^{\frac{3}{2}}}{1215x^5}+C$$

Commented by jagoll last updated on 01/Jan/20

$$sorrysir.theequation\int \frac{\sqrt{9-x^2}}{x^6}dx.$$

Commented by MJS last updated on 01/Jan/20

$$\mathrm{thank}\mathrm{you},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{a}\mathrm{typo},\mathrm{I}\mathrm{corrected}\mathrm{it}$$

Commented by jagoll last updated on 01/Jan/20

$$thanksyousir$$

Answered by petrochengula last updated on 03/Jan/20

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

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