Prove that the integral of $\frac{dx}{\sqrt{x-x^2}}$ is equal to $2\sin^{-1}\sqrt x+C$ using $u = \sqrt{x}$.

Question

I know that you have to change $u=\sqrt x $ to $u^2 = x$. But I don't know what to do next?

Answer 1

As was suggested in your hint, write $$\int \frac{dx}{\sqrt{x-x^2}}=\int \frac{dx}{\sqrt{x}\cdot\sqrt{1-x}}.$$ Now substitute $u=\sqrt x$ to obtan $du=\frac{dx}{2\sqrt x}$ from where you get $$\int \frac{dx}{\sqrt{x-x^2}}=\int \frac{2du}{\sqrt{1-u^2}}=2\arcsin u+C=2\arcsin \sqrt x+C.$$

Answer 2

Hint:

From $u^2=x$ we have $2udu=dx$ and: $$ \int \frac{dx}{\sqrt{x-x^2}}=\int \frac{2udu}{\sqrt{u^2-u^4}}=\int \frac{2udu}{|u|\sqrt{1-u^2}} $$

can you do from this?