# Evaluate the indefinite integral integrate of ((dx)/(x^2(sqrt(9-x^2))))

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Evaluate `int (dx)/(x^2sqrt(9-x^2))` :

Let `x=3sintheta` Then `dx=3costheta d theta;sqrt(9-x^2)=3costheta;x^2=9sin^2theta`

** `sqrt(9-x^2)=sqrt(9-(3sintheta)^2)=3sqrt(1-sin^2theta)=3costheta` **

So `int (dx)/(x^2sqrt(9-x^2))`

`=int (3costheta d theta)/((9sin^2theta)(3costheta))`

`=1/9 int (d theta)/(sin^2 theta)`

`=1/9 int csc^2 theta d theta`

`=-1/9 cottheta +C`

But `x=3sintheta` so `sintheta=x/3` . In using this substitution we imagined a right triangle with acute angle `theta` , side adjacent to `theta` to be `sqrt(9-x^2)` , side opposite `theta` to be `x` , with hypotenuse 3.

Then `cot theta=sqrt(9-x^2)/x`

Now `-1/9 cot theta +C=-1/9(sqrt(9-x^2)/x)+C`

`=-sqrt(9-x^2)/(9x) +C`

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`int (dx)/(x^2sqrt(9-x^2))=-sqrt(9-x^2)/(9x)+C`

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