Question

Evaluate the integral

\(\int {{{\sin }^{ - 1}}xdx} \)

Short answer

We should use the substitution method and integration by parts method to solve the integral.

Step-by-step solution

Given data

Let\(I = \int {{{\sin }^{ - 1}}xdx} \) 🡪(1)

By using integration by parts

\(I = uv - \int {vdu} \)

Let\(u = {\sin ^{ - 1}}x\) \(dv = dx\)

\(du = \frac{1}{{\sqrt {1 + {x^2}} }}dx\) \(v = x\)

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

Substitute the values

Let\(w = 1 + {x^2}\)

\(dw = - 2xdx\)

\(xdx = \frac{{ - dw}}{2}\)

Substitutethe above equation in (2)

\(I = x{\sin ^{ - 1}}x - \int {\frac{1}{{\sqrt w }}\left( {\frac{{ - dw}}{2}} \right)} \)

\(I = x{\sin ^{ - 1}}x + \frac{1}{2}\int {\frac{{dw}}{{\sqrt w }}} \)

\(I = x{\sin ^{ - 1}}x + \frac{1}{x}\left( {2\sqrt w } \right)\)

\(I = x{\sin ^{ - 1}}x + \sqrt w \) 🡪(3)

Reassigning values

\(I = x{\sin ^{ - 1}}x + \sqrt {1 + {x^2}} + c\) where c is constant

Therefore, the integral value of \(\int {{{\sin }^{ - 1}}xdx} \) is \(x{\sin ^{ - 1}}x + \sqrt {1 + {x^2}} + c\)