Evaluate the integral from 0 to 12 of cos(1x) dx

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