Question

The value of $\displaystyle \int_{0}^{\pi/6} \sin 3x \, dx$ is:

• A. $-\frac{\sqrt{3}}{2}$
• B. $-\frac{1}{3}$
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{1}{3}$

Hint:

For $\int \sin(ax) dx$, divide by $a$ and add a negative sign due to the derivative of cosine.

Answer 1

The Correct Option is D

To solve this definite integral, we first find the indefinite integral: \[ \int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C. \] Here, $a = 3$, so: \[ \int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C. \] Now, apply the limits: \[ \int_{0}^{\pi/6} \sin(3x) \, dx = \Big[ -\frac{1}{3} \cos(3x) \Big]_{0}^{\pi/6}. \] Calculate the cosine values: \[ \cos\Big(3 \times \frac{\pi}{6}\Big) = \cos\Big(\frac{\pi}{2}\Big) = 0, \cos(0) = 1. \] So, \[ \int_{0}^{\pi/6} \sin(3x) \, dx = -\frac{1}{3} [\cos(\pi/2) - \cos(0)] = -\frac{1}{3} [0 - 1] = -\frac{1}{3}(-1) = \frac{1}{3}. \] Therefore, the answer is $\frac{1{3}$}.