To evaluate the integral ∫₀¹² cos(1x) dx, we start by finding the antiderivative of cos(1x).
Using the substitution technique, let u = 1x which implies du = dx. The integral now becomes:
∫ cos(u) du
The antiderivative of cos(u) is sin(u) plus the constant of integration C. Thus, we have:
sin(u) + C
Substituting back u = 1x, we get:
sin(1x) + C
Now we evaluate the definite integral from 0 to 12:
∫₀¹² cos(1x) dx = [sin(1x)]₀¹² = sin(12) – sin(0)
Since sin(0) = 0, we find that:
∫₀¹² cos(1x) dx = sin(12) – 0 = sin(12)
So, the final answer for the integral is:
sin(12) + C