evaluate the following integral `int 1/{x^2sqrt{25-x^2}}dx`

please explain as you solve

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To solve this integral, we need to use the trig substitution `x=5sinu`

This means that `dx=5cosudu`

and we can also use some algebra to find

`x^2=25sin^2u`

and `sqrt{25-x^2}=sqrt{25(1-sin^2u)}=5cosu`

using the identity `sin^2u+cos^2u=1` .

This means the integral becomes

`int 1/{x^2sqrt{25-x^2}}dx`

`=int 1/{25sin^2u (5cosu)}5cos udu`

`=1/25 int 1/{sin^2 u}du`

`=1/25 int csc^2udu` but `d/{du} cot u=-csc^2u`

`=-1/25 cot u+C` where C is the constant of integration

`=-1/25 sqrt{25-x^2}/x+C` using the pythagorean theorem with the u triangle

**The integral evaluates to `-1/25 sqrt{25-x^2}/x+C` .**

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