Evaluate $$\int_{0}^{\frac{\pi}{6}} \cos(3x - \frac{\pi}{6}) \, dx.$$ - Scottish Highers Maths - Question 11 - 2022

Now, we complete the integration:

$\int \cos(3x - \frac{\pi}{6}) \, dx = \frac{1}{3} \sin(3x - \frac{\pi}{6}) + C.$

Next, we substitute the limits of integration:

$\left[ \frac{1}{3} \sin(3x - \frac{\pi}{6}) \right]_{0}^{\frac{\pi}{6}}.$

Calculating the first part:
At $x = \frac{\pi}{6}$: $\sin(3 \cdot \frac{\pi}{6} - \frac{\pi}{6}) = \sin(\frac{\pi}{2}) = 1.$
At $x = 0:$
\sin(3 \cdot 0 - \frac{\pi}{6}) = \sin(-\frac{\pi}{6}) = -\frac{1}{2.

Now substituting those into our expression gives: $\frac{1}{3} [1 - (-\frac{1}{2})] = \frac{1}{3} [1 + \frac{1}{2}] = \frac{1}{3} \cdot \frac{3}{2} = \frac{1}{2}.$

Thus, the final answer is:

$\frac{1}{2}.$