integral sin (3x+5)dx batas bawah 0 batas atas phi/12

jawab

π/2
∫  sin(3x + 5) dx
0


= [- 1/3 cos (3x  + 5) ] (90...0)

= - 1/3 [ cos (3(90) + 5) - cos (3(0) + 5)}
= -1/3 [ cos 275 - cos 5]
= - 1/3 [ -  2 sin 1/2 (275+5) sin 1/2 (275- 5)}
= - 1/3 [ - 2 sin 140 sin 135 ]
= - 2/3 [ sin 140 . (1/2 √2]
= - 1/3 √2. sin 140