Question

integral (sin^-1)² .plz solve this

Answer 1

Heya,

∫ (sin⁻¹x)²dx.

let sin⁻¹x = t
x = sint
dx = cost * dt

now,

∫ t² * cost * dt

Integration by parts, taking t² as fisrt function,

We get,

t²∫cost * dt - ∫[dt²/dt * ∫cost * dt ] dt

t²(sint) - ∫2t(sint) * dt

t²(sint) - 2[ t(-cost) - ∫1*(-cost)dt]
[ again integrating by parts, taking t as first function]

t²(sint) + 2tcost - 2sint + c

now put the value if t in the Equation.

x(sin⁻¹x)² + 2(sin⁻¹)√(1 - x²) - 2x + C
where, c is constant

hope this help you....,