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

Dec 21, 2016

$- \left(\sqrt{1 - {x}^{2}}\right) + C$

#### Explanation:

Using

${\sin}^{2} x + {\cos}^{2} x = 1 \therefore {\sin}^{2} x = 1 - {\cos}^{2} x$

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

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

we have:$\text{ } \int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx} = \int \frac{\sin u}{\sqrt{1 - {\sin}^{2} u}} \times \left(\cos u\right) \mathrm{du}$

$\text{ } = \int \frac{\sin u}{\cancel{\cos} u} \times \cancel{\left(\cos u\right)} \mathrm{du}$

$= \int \sin u \mathrm{du} = - \cos u + C$

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

this can also be integrated by inspection

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

we note that a function of the derivative is outside the bracket ,so:

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

result follows

Dec 21, 2016

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

#### Explanation:

The integrand is defined only for $x \in \left(- 1 , 1\right)$ so we can substitute $x = \sin t$, $\mathrm{dx} = \cos t \mathrm{dt}$ with $t \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$

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

In the same interval,

$\cos t = \sqrt{1 - {x}^{2}}$

So that:

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