Nilai dari Integral 4sin7xcos3x batas atas π/6 batas bawah 0

Penyelesaian ₀∫^π/64sin7xcos3x dx dengan menggunakan rumus:

2 sinα cosβ = sin(α + β) + sin(α - β)

Jawab:

₀∫^π/64 sin 7x cos 3x dx

= 2 ₀∫^π/6sin(7x + 3x) + sin (7x - 3x)dx

= 2 ₀∫^π/6 (sin 10x + sin 4x) dx

= 2 [{-1 cos 10x - 1 cos 4x}]₀^π/6
10 4

= (-1 cos 10π - 1 cos 4π) - (-1 cos 0 - 1 cos 0)

5 6 2 6 5 2

= (-1 cos 300° - 1 cos 120° ) - (-1 . 1 - 1 . 1)

5 2 5 2

= (- 1 . 1 - 1 (- 1) + 1 + 1

5 2 2 2 5 2

= (- 4 + 10 + 8 + 20)

40 40 40 40

= 34

= 17