Evaluate \(\int \sin 3x \mathrm d x\)

  • A (2/3) cos 3x + c
  • B (1/3) cos 3x + c
  • C (-1/3) cos 3x + c
  • D (-2/3) cos 3x + c
jamb 2002mathematics

The correct answer is C. (-1/3) cos 3x + c

To evaluate the integral \(\int \sin 3x \, dx\), we'll use the integral formula for the sine function:

\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C,\)

where \(a\) is a constant and \(C\) is the constant of integration.

In this case, \(a = 3\), so we have:

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

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