Question

Using \[t=1-x^2,\] what is \[\int \frac{x\,dx}{\sqrt{1-x^2}}?\]

पर्याय

• \[-\sqrt{1-x^2}\]

• \[\sqrt{1-x^2}\]

• \[\frac{1}{\sqrt{1-x^2}}\]

• \[-\frac{1}{\sqrt{1-x^2}}\]

Answer 1

With \(t=1-x^2\), \(dt=-2x\,dx\). Therefore, the integral becomes \(-\frac12\int \frac{dt}{\sqrt t}=-\sqrt t=-\sqrt{1-x^2}\).