# How do you integrate int 1/(x^2sqrt(9-x^2))dx using trigonometric substitution?

Apr 6, 2018

$\int \frac{1}{{x}^{2} \sqrt{9 - {x}^{2}}} \mathrm{dx} = - \frac{1}{9} \left(\frac{\sqrt{9 - {x}^{2}}}{x}\right) + c$

#### Explanation:

${I}_{1} = \int \frac{1}{{x}^{2} \sqrt{9 - {x}^{2}}} \mathrm{dx}$

$x = 3 \sin u \implies \mathrm{dx} = 3 \cos u \mathrm{du}$

${I}_{1} = \int \frac{1}{{\left(3 \sin u\right)}^{2} \sqrt{9 - 9 {\sin}^{2} u}} 3 \cos u \mathrm{du}$

${I}_{1} = \int \frac{1}{\left(9 {\sin}^{2} u\right) \cancel{3 \sqrt{1 - {\sin}^{2} u}}} \cancel{3 \cos u} \mathrm{du}$

${I}_{1} - \frac{1}{9.} {\sin}^{2} u \mathrm{du} = \frac{1}{9} \int {\csc}^{2} u \mathrm{du}$

this si a satndard integral

${I}_{1} = - \frac{1}{9} \cot u + c$

$\because \sin u = \frac{x}{3} , \cos u = \frac{\sqrt{9 - {x}^{2}}}{3}$

by Pythagoras, this is left for the reader to verify

$\therefore - \frac{1}{9} \cot u = - \frac{1}{9} \frac{\frac{\sqrt{9 - {x}^{2}}}{3}}{\frac{x}{3}} + c$

$= - \frac{1}{9} \left(\frac{\sqrt{9 - {x}^{2}}}{x}\right) + c$

$\int \frac{1}{{x}^{2} \sqrt{9 - {x}^{2}}} \mathrm{dx} = - \frac{1}{9} \left(\frac{\sqrt{9 - {x}^{2}}}{x}\right) + c$