Question

Soal Kuis
Nomor 1 :


Nomor 2 :

Answer 1

∫ sin(3x) + cos(3x) dx
= 1/3 (sin(3x) - cos(3x), masukkan batasan:
= [1/3 (sin(3(30)) - cos(3(30))] - [1/3 (sin(3(0)) - cos(3(0))]
= [1/3 (sin(90) - cos(90))] - [1/3 (sin(0) - cos(0))]
= [1/3 (1 - 0)] - [1/3 (0 - 1)]
= [1/3 (1)] - [1/3 (-1)]
= 1/3 - (-1/3)
= 1/3 + 1/3
= 2/3

∫ tan²(x) dx
= tan(x) - x, masukkan batasan
= [tan(x) - x] - [tan(x) - x]
= [tan(45) - 45°] - [tan(0) - 0]
= [1 - 45°] - [0 - 0]
= (1 - 45°) - 0
= 1 - 45°
= 1 - π/4

Answer 2

∫ sin(3x) + cos(3x) dx
= 1/3 (sin(3x) - cos(3x), {³°₋°}
= [1/3 (sin(3(30)) - cos(3(30))] - [1/3 (sin(3(0)) - cos(3(0))]
= [1/3 (sin(90) - cos(90))] - [1/3 (sin(0) - cos(0))]
= [1/3 (1 - 0)] - [1/3 (0 - 1)]
= [1/3 (1)] - [1/3 (-1)]
= 1/3 - (-1/3)
= 1/3 + 1/3
= 2/3

∫ tan²(x) dx
= tan(x) - x,
= [tan(x) - x] - [tan(x) - x]
= [tan(45) - phi/4] - [tan(0) - 0]
= [1 - phi/4] - [0 - 0]
= 1-phi/4