Find the definite integration of sin(4x+π/6) dx which has π/6 and 0 as limits

Answer 1

Using the formula for integration of sin(ax+b), we have:

∫sin(4x+π/6) dx = -1/4 cos(4x+π/6) + C

where C is the constant of integration.

Evaluating the definite integral with limits π/6 and 0:

∫sin(4x+π/6) dx [π/6,0] = [-1/4 cos(4(0)+π/6)] - [-1/4 cos(4(π/6)+π/6)]

= [-1/4 cos(π/6)] - [-1/4 cos(5π/6)]

= [-1/4 (sqrt(3)/2)] - [-1/4 (-sqrt(3)/2)]

= -sqrt(3)/8 + sqrt(3)/8

= 0

Therefore, the definite integration of sin(4x+π/6) dx with limits π/6 and 0 is 0.