Question

Write the integral as the sum of the integral of an odd function and the integral of an even function. Use this simplification to evaluate the integral. $$ \int_{-\pi}^{\pi}(\sin 3 x+\cos 3 x) d x $$

Short answer

The integral of the given function from -π to π is 0.

Step-by-step solution

Identify the type of each function.

The integral contains two functions: \(\sin 3x\) and \(\cos 3x\). As mentioned above, \(\sin x\) is an odd function and \(\cos x\) is an even function. These properties remain the same when the functions become \(\sin 3x\) and \(\cos 3x\).

The integral of an odd function.

Know that the integral of an odd function over a symmetric interval around zero (like -\(\pi\) to \(\pi\)) is always zero. In this case, \(\sin 3x\) is an odd function. Therefore, the integral of \(\sin 3x\) from -\(\pi\) to \(\pi\) is zero.

The integral of an even function.

The integral of an even function over a symmetric interval around zero can be simplified as twice the integral from 0 to the upper boundary of the original interval. So, the integral of \(\cos 3x\) from -\(\pi\) to \(\pi\) is twice the integral of \(\cos 3x\) from 0 to \(\pi\). Integrating, we get \(2 \int_{0}^{\pi} \cos 3x dx=2[\frac{\sin3x}{3}]_0^\pi = 2 * (\frac{\sin3\pi}{3}-\frac{\sin0}{3}) \).

Evaluate the integral of the even function.

Evaluate the expression \(2 * (\frac{\sin3\pi}{3}-\frac{\sin0}{3})\). Now, we know that \(\sin3\pi = \sin(2\pi + \pi) = \sin0 = 0\). So, our expression becomes \(2*(0 - 0) = 0\)