nilai integral batas atas ⅓π batas bawah 0 (sin 2x + 3 cos x)dx=

InTegraL

∫(sin 2x + 3cos x) dx [1/3 π .. 0]
= ∫sin2x dx + 3 ∫cos x dx
= (-1/2) ∫dcos 2x + 3 ∫dsin x
= -1/2 cos 2x + 3 sin x
= (-1/2 cos (2.1/3 π) + 3 sin 1/3 π) - (-1/2 cos 0 + 3 sin 0)
= (-1/2 cos 120° + 3 sin 60°) - (-1/2 . 1 + 3 . 0)
= (-1/2 . -1/2 + 3 . 1/2 √3) + 1/2
= 1/4 + (3/2)√3 + 1/2
= 3/4 + (3/2)√3
= (3/4)(1 + 2√3)