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Evaluate the following:\( \int \frac{\sqrt{x}}{\sqrt{1-x^{2}}} d x=? \) |
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Answer» \(\int\frac{\sqrt x}{\sqrt{1-x^2}}dx\) 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 |
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