Question

Tolong jelaskan caranya, masih gak paham
integral x^2 / akar (9 - x^2)

Answer 1

Misal x = 3 sin t , maka
sin t= x/3
cos t = ¹/₃ √(9-x²)
x^2  = 9 sin^2 t
dx =  3 cos t dt
t = sin⁻¹ (x/3)

∫ x² / √(9-x²) dx = ∫ (9 sin² t. 3 cos t dt) /(cos t )
= ∫ 9 sin² t dt = ∫ 9 (¹/₂-  ¹/₂ cos 2t ) dt = ⁹/₂ ∫  (1 - cos 2t) dx
= ⁹/₂ ( t - ¹/₂ sin 2 t ) +c
= ⁹/₂ ( t - ¹/₂ . 2 sin t cos t ) +c
= ⁹/₂ ( t - sin t cos t) +c
= ⁹/₂ {(sin⁻¹(x/3) - (ˣ/₃)(¹/₃ √(9-x²)} + c
= ⁹/₂ { sin⁻¹(x/3) - ˣ/₉ √(9-x²)} + c
= ¹/₂ ( 9 sin¹(ˣ/₃) - x√(9-x²) + c