How is it that $-1/2$ can be factored outside of $\int \frac{x}{\sqrt {1-x^2}}dx$?

Question

If you are given

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

How is it that $-\frac{1}{2}$ can be factored outside of the integral?

The exponent in the denominator is $\frac{1}{2}$, but I do not see where 2 is in the denominator.

Answer 1

You can write $$-\frac{1}{2}\int\frac{-2x}{\sqrt{1-x^2}}dx$$ since $$(1-x^2)'=-2x$$

Answer 2

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

Let $1-x^2=u$ => Differentiate with respect to 'x'

$dx=\frac{du}{-2x}$

Now, $\int \frac{x}{\sqrt {u}}$$\frac{du}{-2x}$

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

=$-\sqrt {u}+C$

Substituting u=$\sqrt {1-x^2}$, then we get

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

Answer 3

I'm not sure if I'm understanding your question right, but in order to factor out a $-\frac 12$, you need to multiply by $-2$ because $-2\times-\frac 12=1$.$$\begin{align*}\int\frac {x}{\sqrt{1-x^2}} & =1\cdot\int\frac x{\sqrt{1-x^2}}\\ & =\frac {-2}{-2}\int\frac x{\sqrt{1-x^2}}\\ & =\frac 1{-2}\int\frac {-2x}{\sqrt{1-x^2}}\\ & =-\frac 12\int\frac {-2x}{\sqrt{1-x^2}}\end{align*}$$As for why we would do that, it's because the derivative of $1-x^2$ is $-2x$ which is the numerator (i.e integrating by u-substitution)