Question

x² sin⁻¹x ,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

HELLO DEAR,

integrating by parts, taking sin^{-1}x as the first function we get

$\int$x²sin^{-1}x.dx = (sin^{-1}) * x³/3 - $\int$1/√(1 - x²) * x³/3.dx

= x³sin^{-1}x/3 - 1/3$\int$x³/√(1 - x²).dx

= x³sin^{-1}x/3 - 1/3$\int$(x.x²)/√(1 - x²).dx

= x³sin^{-1}x/3 + 1/3$\int${t(1 - t²)/t.dt

where, (1 - x²) = t²

= x³sin^{-1}x/3 + 1/3$\int$dt - 1/3$\int$t².dt

= x³sin^{-1}x /3 + 1/3t - 1/9t³ + c

= x³sin^{-1}x /3 + 1/3(√1 - x²) - 1/8(1 - x)³/² + c.

I HOPE ITS HELP YOU DEAR,

THANKS