Question

Find:  integrate e ^{log sin^-1x}  dx.

 \(\int\)\(e^{logsin^{-1} x}\)dx

Answer 1

\(\int\)\(e^{logsin^{-1} x}\)dx = \(\int\)sin^-1x dx (\(\because\)e^logf(x) = f(x))

= sin^-1x\(\int1dx\) - \(\int(\frac{d}{dx}sin^{-1}x(\int 1dx))dx\) 

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

 = x sin^-1x  + \(\frac12\)\(\int\frac{-2x}{\sqrt{1-x^2}}dx\)

Let 1 - x² = t²

⇒ -2x dt = 2tdt

= x sin^-1x + \(\frac12\int\frac{2tdt}t \)

 = x sin^-1x + t + c, where c is integral constant

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