# How do you find the integral of 1/(x^2sqrt(25-x^2))dx?

Mar 1, 2015

The answer is: $- \frac{\sqrt{25 - {x}^{2}}}{25 x} + c$.

We have make a substitution:

$x = 5 \sin t \Rightarrow \mathrm{dx} = 5 \cos t \mathrm{dt}$,

so:

$\int \frac{1}{{x}^{2} \sqrt{25 - {x}^{2}}} \mathrm{dx} = \int \frac{1}{{\left(5 \sin t\right)}^{2} \sqrt{25 - 25 {\sin}^{2} t}} \cdot 5 \cos t \mathrm{dt} =$

$= \int \frac{5 \cos t}{{\left(5 \sin t\right)}^{2} \sqrt{25 \left(1 - {\sin}^{2} t\right)}} \mathrm{dt} =$

$= \int \frac{5 \cos t}{25 {\sin}^{2} t \cdot 5 \cos t} \mathrm{dt} = \frac{1}{25} \int \frac{1}{\sin} ^ 2 t \mathrm{dt} =$

$= - \frac{1}{25} \int - \frac{1}{\sin} ^ 2 t \mathrm{dt} = - \frac{1}{25} \cot t + c = - \frac{1}{25} \cos \frac{t}{\sin} t + c = \left(1\right)$

since

$\sin t = \frac{x}{5}$ and $\cos t = \sqrt{1 - {\sin}^{2} t} = \sqrt{1 - {x}^{2} / 25} = \frac{\sqrt{25 - {x}^{2}}}{5}$

than:

$\left(1\right) = - \frac{1}{25} \frac{\sqrt{25 - {x}^{2}}}{5} \cdot \frac{5}{x} + c = - \frac{\sqrt{25 - {x}^{2}}}{25 x} + c$.