Evaluate the integral

Question

∫x⁵√(1-x²)dx

Answer 1

∫ x⁵√(1 - x²)dx = ∫x⁴√(1-x²)xdx

Let u = 1-x². Then du = -2xdx. So, xdx = (-1/2)du. Also, x² = 1-u, so x⁴ = (1-u)².

We have (-1/2)∫ (1-u)²√udu = (-1/2) ∫ (1 - 2u + u²)√udu

= (-1/2) ∫ (u^1/2 - 2u^3/2 + u^5/2)du = (-1/2) [(2/3)u^3/2 - (4/5)u^5/2 + (2/7)u^7/2] + C

= (-1/3)(1-x²)^3/2 + (2/5)(1 - x²)^5/2 - (1/7)(1 - x²)^7/2 + C