Find the definite integration of sin(4x+π/6) dx which has π/6 and 0 as limits
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.
To find the definite integral of sin(4x+π/6) with limits π/6 and 0, follow these steps:
Step 1: Start with the given function: sin(4x+π/6).
Step 2: Apply the integration formula for the sine function, which is -cos(4x+π/6)/4.
Step 3: Evaluate the definite integral by substituting the upper limit (π/6) into the antiderivative and subtracting the result from substituting the lower limit (0) into the antiderivative.
∫[0 to π/6] sin(4x+π/6) dx = [-cos(4(π/6)+π/6)/4] - [-cos(4(0)+π/6)/4]
Step 4: Simplify the expression:
= [-cos(2π/3+π/6)/4] - [-cos(0+π/6)/4]
= [-cos(5π/6)/4] - [-cos(π/6)/4]
Step 5: Simplify further by computing the cosine values:
= [-(-√3/2)/4] - [-(√3/2)/4]
= [√3/8] - [√3/8]
Step 6: Subtract the two terms together to get the final result:
= √3/8 - √3/8
= 0
Therefore, the definite integral of sin(4x+π/6) with limits π/6 and 0 is equal to zero.