MHB How can we evaluate this indefinite integral of a definite integral?

[juantheron]

Evaluation of Indefinite Integral $\displaystyle \int_{0}^{1} \sqrt{1-2\sqrt{x-x^2}}dx$

$\bf{My\; Try::}$ We can write the given Integral as

$\displaystyle \int_{0}^{1}\sqrt{\left(\sqrt{x}\right)^2+\left(\sqrt{1-x}\right)^2-2\sqrt{x}\cdot \sqrt{1-x}}dx$

So Integral Convert into $\displaystyle \int_{0}^{1}\left|\sqrt{x}-\sqrt{1-x}\right|dx$

Now How can i solve after that , explanation Required.

Thanks

 

[chisigma]

Evaluation of Indefinite Integral $\displaystyle \int_{0}^{1} \sqrt{1-2\sqrt{x-x^2}}dx$

$\bf{My\; Try::}$ We can write the given Integral as

$\displaystyle \int_{0}^{1}\sqrt{\left(\sqrt{x}\right)^2+\left(\sqrt{1-x}\right)^2-2\sqrt{x}\cdot \sqrt{1-x}}dx$

So Integral Convert into $\displaystyle \int_{0}^{1}\left|\sqrt{x}-\sqrt{1-x}\right|dx$

Now How can i solve after that , explanation Required.

Thanks

... the successive step is easy...

$\displaystyle \int_{0}^{1} |\sqrt{x} - \sqrt{1 - x}|\ dx = \int_{0}^{\frac{1}{2}} (\sqrt{1 - x} - \sqrt{x})\ dx + \int_{\frac{1}{2}}^{1} (\sqrt{x} - \sqrt{1 - x})\ d x $

Kind regards

$\chi$ $\sigma$

 

[Prove It]

This is actually a DEFINITE integral...