Question

\(6 \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x)dx\) is equal to:

Hint:

A useful property for sine integrals over \([0, \pi]\) is \(\int_0^\pi \sin(nx) dx\).
If `n` is an even integer, the integral is 0. If `n` is an odd integer, the integral is \(2/n\).
Using this: \(\int_0^\pi \sin(3x)dx = 2/3\), \(\int_0^\pi \sin(2x)dx = 0\), \(\int_0^\pi \sin(x)dx = 2\).
The sum is \(2/3 + 0 + 2 = 8/3\). Multiplying by 6 gives 16. This shortcut is much faster.

Answer 1

Correct Answer: 40

Step 1: Understanding the Question:
We are required to evaluate a definite integral involving a sum of sine functions over the interval from \(0\) to \(\pi\), and then multiply the result by 6.

Step 2: Formulae Used:
The required integration rule is: \[ \int \sin(ax)\,dx = -\frac{1}{a}\cos(ax) + C \] Each term in the integrand will be integrated separately.

Step 3: Detailed Solution:
Let \[ I = \int_{0}^{\pi} (\sin 3x + \sin 2x + \sin x)\,dx \] Integrating term by term: \[ I = \left[ -\frac{\cos(3x)}{3} - \frac{\cos(2x)}{2} - \cos x \right]_{0}^{\pi} \] Evaluating at the upper limit \(x = \pi\): \[ -\frac{\cos(3\pi)}{3} - \frac{\cos(2\pi)}{2} - \cos(\pi) = \frac{1}{3} - \frac{1}{2} + 1 \] Evaluating at the lower limit \(x = 0\): \[ -\frac{\cos(0)}{3} - \frac{\cos(0)}{2} - \cos(0) = -\frac{1}{3} - \frac{1}{2} - 1 \] Subtracting: \[ I = \left(\frac{1}{3} - \frac{1}{2} + 1\right) - \left(-\frac{1}{3} - \frac{1}{2} - 1\right) \] \[ I = \frac{2}{3} + 2 = \frac{8}{3} \] Now multiplying by 6: \[ 6I = 6 \times \frac{8}{3} = 16 \] Step 4: Final Answer:
\[ \boxed{16} \]