How to Integrate sqrt(x/2-x)dx Using Trigonometric Substitution?

[DigiDigi]

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

 

[junaid314159]

Is the problem:

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

 

[Infrared]

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.

 

[DigiDigi]

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]

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

 

[DigiDigi]

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