Soal Integral 2

Contoh 1:

∫ sin³ 2x dx = ....

Jawab:
∫ sin³ 2x dx

= ∫ (sin² 2x)(sin 2x) dx

=  ∫ ½(1– cos 4x)(sin 2x) dx; ingat: cos 4x = 1 – 2 sin² 2x

= ½ ∫ (sin 2x – cos 4x sin 2x) dx

= ½ ∫ sin 2x dx – ½ ∫ (cos 4x sin 2x) dx

= ½ ∫ sin 2x dx – ½ ∫ ½{sin (4x + 2x) – sin (4x – 2x)} dx

= ½ ∫ sin 2x dx – ¼ ∫ sin 6x dx – ¼ ∫ sin 2x dx

= –¼ cos 2x + (1/24) cos 6x + ⅛ cos 2x + c

= – ⅛ cos 2x + (1/24) cos 6x + c

RUMUS CEPATNYA ADA DI NICEinstitute

Contoh 2:

Jika dy/dx = (x + 3x²) y² dan y = 6 saat x = 0, maka y = ....


Jawab:
dy/dx = (x + 3x²) y²

y²(dy/dx) = (x + 3x²) 

y²dy = (x + 3x²) dx
∫ y²dy = ∫ (x + 3x²) dx

⅓ y³ + C₁ = ½x² + x³ + C₂ 

y³ + 3C₁ = (3/2)x² + 3x³ + 3C₂

y³ = (3/2)x² + 3x³ + 3C₂ – 3C₁

y³ = (3/2)x² + 3x³ + C

216 = (3/2)(0)² + 3(0)³ + C

C = 216

y = ∛{(3/2)x² + 3x³ + C}