\( \int \frac{\sqrt{x}}{\sqrt{1-x^{2}}} d x=? \)

Question

Please log in or register to answer this question.

ShlokShukla · Answer 1 · 2022-05-10T07:23:45+0000

\(\int\frac{\sqrt x}{\sqrt{1-x^2}}dx\)

Let

1 - x² = t²

-2xdx = 2tdt

⇒ xdx = -tdt

∴ \(\int\frac{\sqrt x\,dx}{\sqrt{1-x^2}}=-\int\frac{tdt}{t\sqrt{1-t^2}}\)

\(= -\int\frac{dt}{\sqrt{1-t^2}}\)

= - sin^-1 t + C

= - sin^-1 \(\sqrt{1-x^2}\) + C