Questions: Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integral [ int0^pi / 6 tan (3 theta) d theta ] Select the correct choice below and fill in the answer box to complete your choice. (Type an exact answer.) A. The integral converges because (int0^pi / 6 tan (3 theta) d theta=) . B. The integral diverges because (int0^pi / 6 tan (3 theta) mathrmd theta=) .

Solution Steps

To solve the integral \(\int_{0}^{\pi / 6} \tan (3 \theta) d \theta\), we can directly integrate the function. The antiderivative of \(\tan(3\theta)\) can be found using a substitution method. Let \(u = 3\theta\), then \(du = 3d\theta\) or \(d\theta = \frac{1}{3}du\). The integral becomes \(\frac{1}{3} \int \tan(u) du\), which can be solved using the known antiderivative of \(\tan(u)\), which is \(-\ln|\cos(u)|\). After finding the antiderivative, evaluate it at the bounds \(\theta = 0\) and \(\theta = \pi/6\).

We need to evaluate the integral

\[ \int_{0}^{\pi / 6} \tan(3\theta) \, d\theta. \]

Using the substitution \(u = 3\theta\), we have \(du = 3 \, d\theta\) or \(d\theta = \frac{1}{3} du\). The limits of integration change as follows:

• When \(\theta = 0\), \(u = 0\).
• When \(\theta = \frac{\pi}{6}\), \(u = \frac{\pi}{2}\).

Thus, the integral becomes:

\[ \int_{0}^{\pi / 6} \tan(3\theta) \, d\theta = \frac{1}{3} \int_{0}^{\frac{\pi}{2}} \tan(u) \, du. \]

The integral of \(\tan(u)\) is

\[ -\ln|\cos(u)|. \]

Evaluating this from \(0\) to \(\frac{\pi}{2}\):

\[ \frac{1}{3} \left[-\ln|\cos(u)| \bigg|_{0}^{\frac{\pi}{2}}\right]. \]

At \(u = 0\):

\[ -\ln|\cos(0)| = -\ln(1) = 0. \]

At \(u = \frac{\pi}{2}\):

\[ -\ln|\cos(\frac{\pi}{2})| = -\ln(0) = \infty. \]

Since the integral diverges to infinity, we conclude that:

\[ \int_{0}^{\pi / 6} \tan(3\theta) \, d\theta = \infty. \]

Final Answer