Question

Evaluate:

`int_0^1 x * sin^-1x * dx`

Answer 1

Let, u = sin⁻¹ x, dv = x dx

Then,

`du = 1/(sqrt(1 - x^2)) dx, v = x^2/2`

So,

`int x * sin^-1 x * dx = x^2/2 sin^-1 x - int x^2/(2sqrt(1 - x^2)) dx`

Evaluate the definite integral:

First term:

`[x^2/2 sin^-1 x]_0^1 = 1/2 * π/2 = π/4`

Second integral:

`int_0^1 x^2/(2sqrt(1 - x^2)) dx`

Put x = sin ⁡θ, then it becomes:

`1/2 int_0^(π/2) sin^2 ⁡θ  d⁡θ = 1/2 * ⁡π/4 = π/8`

Hence,

`π/4 - π/8`

= `π/8`