Question

evaluate ∫sin−1x dx

Answer 1

We need to evaluate ∫ sin^-1x dx

Let u = sin^-1x, dv = dx

⇒ du = \(\frac{1}{\sqrt{1-x^2}}\)dx, v = x

∫ sin^-1x dx = x sin^-1x - ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx + c ....(1)

where c is integration constant.

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

Let w = 1 - x²

⇒ dw = -2xdx

⇒ xdx = -dw/2

∫ \(\frac{x}{\sqrt{1-x^2}}\)dx = ∫ \(-\frac{dw}{2\sqrt w}\)

= -1/2 ∫ dw/√w

= -1/2 (2√w)

= -√w

substitute back w = 1 - x² we get

∴ ∫ \(\frac{x}{\sqrt{1-x^2}}\)dx = \(-\sqrt{1-x^2}\)

Therefore (1) becomes

∫ sin^-1x dx = x sin^-1 x + \(\sqrt{1-x^2}\) + c