# Integrate sqrt(x/2-x)dx

## Main Question or Discussion Point

How do I integrate sqrt(x/2-x)dx?

Is the problem:

$\int\sqrt{\frac{x}{2}-x}\space dx$ or $\int\sqrt \frac{x}{2-x}\space dx$ ?

Infrared
Gold Member
I'm assuming it is the later. $\int \sqrt{\frac{x}{2-x}} dx= \int \frac{\sqrt{x}}{\sqrt{2-x}} dx$

Use the substitution $u= \sqrt{2-x}$ to turn the integral into $-2 \int \sqrt{2-u^2} du$, which you can do by trig substitution.

I'm assuming it is the later. $\int \sqrt{\frac{x}{2-x}} dx= \int \frac{\sqrt{x}}{\sqrt{2-x}} dx$

Use the substitution $u= \sqrt{2-x}$ to turn the integral into $-2 \int \sqrt{2-u^2} du$, which you can do by trig substitution.
Yes,it this one.I know U substitution, but I think I miss out something.How did you change $u= \sqrt{2-x}$ to turn the integral into $-2 \int \sqrt{2-u^2} du$

Infrared
Gold Member
If $u=\sqrt{2-x}$, then $\sqrt{x}=\sqrt{2-u^2}$ and $du=-\frac{dx}{2\sqrt{2-x}}$.

When you said trigonometric substitution, you mean a^2-x^2=1-sin^2(x)?

Infrared
More like the $a^2-x^2=a^2(1-sin^2{\theta})$ that you will get when you set $u^2=2sin^2\theta$