Question

Evaluate \[\int \frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.\]

विकल्प

• \[x+\sqrt{1-x^2}\sin^{-1}x+C\]

• \[-x-\sqrt{1-x^2}\sin^{-1}x+C\]

• \[x-\sqrt{1-x^2}\sin^{-1}x+C\]

• \[\sqrt{1-x^2}\sin^{-1}x-x+C\]

Answer 1

Use \(u=\sin^{-1}x\) and \(v=-\sqrt{1-x^2}\). Then \(\int u\,dv=-\sqrt{1-x^2}\sin^{-1}x+x+C\), which is the stated result.