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

Jan 27, 2017

You don't. By inspection, the answer is $\left(- \frac{1}{2}\right) \sqrt{1 - {x}^{2}} + c$

#### Explanation:

It's of the form $\frac{f ' \left(x\right)}{\sqrt{f \left(x\right)}}$. In other words, the thing on the top is (nearly) the derivative of the thing under the square root. When you differentiate sqrt(f(x) you get $\left(\frac{1}{2}\right) \frac{f ' \left(x\right)}{\sqrt{f \left(x\right)}}$ (chain rule).

Jan 27, 2017

$\int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = - \sqrt{1 - {x}^{2}} + C$

#### Explanation:

By substitution:

Let $u = 1 - {x}^{2}$. Then $\mathrm{du} = - 2 x \mathrm{dx}$. Substituting, we get

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

$= - \frac{1}{2} \int \frac{1}{\sqrt{u}} \mathrm{du}$

$= - \frac{1}{2} \int {u}^{- \frac{1}{2}} \mathrm{du}$

$= - \frac{1}{2} \left(\frac{{u}^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}\right) + C$

$= - \frac{1}{2} \left(\frac{{u}^{\frac{1}{2}}}{\frac{1}{2}}\right) + C$

$= - \sqrt{u} + C$

$= - \sqrt{1 - {x}^{2}} + C$