integral (sin^-1)² .plz solve this
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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....,
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