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How do I integrate sqrt(x/2-x)dx?
Yes,it this one.I know U substitution, but I think I miss out something.How did you change [itex] u= \sqrt{2-x} [/itex] to turn the integral into [itex] -2 \int \sqrt{2-u^2} du [/itex]I'm assuming it is the later. [itex] \int \sqrt{\frac{x}{2-x}} dx= \int \frac{\sqrt{x}}{\sqrt{2-x}} dx[/itex]
Use the substitution [itex] u= \sqrt{2-x} [/itex] to turn the integral into [itex] -2 \int \sqrt{2-u^2} du [/itex], which you can do by trig substitution.