Tolong jelaskan caranya, masih gak paham integral x^2 / akar (9 - x^2)
DB45
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
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