use integration, the direct comparison test, or the limit comparison test to test the integral for…

use integration, the direct comparison test, or the limit comparison test to test the integral for convergence. if more than one method applies, use whatever method you pre\n\nselect the correct choice below and fill in the answer box to complete your choice.\n(type an exact answer.)\n\na. the integral converges because (int_{0}^{pi / 6} \tan (3 \theta) d \theta=)\n\nb. the integral diverges because (int_{0}^{pi / 6} \tan (3 \theta) d \theta=)

use integration, the direct comparison test, or the limit comparison test to test the integral for convergence. if more than one method applies, use whatever method you pre\n\nselect the correct choice below and fill in the answer box to complete your choice.\n(type an exact answer.)\n\na. the integral converges because (int_{0}^{pi / 6} \tan (3 \theta) d \theta=)\n\nb. the integral diverges because (int_{0}^{pi / 6} \tan (3 \theta) d \theta=)

Answer

Explanation:

Step1: Integrate (\tan(3\theta))

Recall that (\int\tan(x)dx=-\ln|\cos x| + C). Let (u = 3\theta), then (du=3d\theta) and (d\theta=\frac{1}{3}du). (\int\tan(3\theta)d\theta=\frac{1}{3}\int\tan(u)du=-\frac{1}{3}\ln|\cos u|+C=-\frac{1}{3}\ln|\cos(3\theta)|+C)

Step2: Evaluate the definite - integral

Using the fundamental theorem of calculus (\int_{a}^{b}f(x)dx=F(b)-F(a)), where (F(x)) is an antiderivative of (f(x)). (\int_{0}^{\frac{\pi}{6}}\tan(3\theta)d\theta=\left[-\frac{1}{3}\ln|\cos(3\theta)|\right]{0}^{\frac{\pi}{6}}) Substitute the upper and lower limits: When (\theta=\frac{\pi}{6}), (\cos\left(3\times\frac{\pi}{6}\right)=\cos\left(\frac{\pi}{2}\right) = 0). When (\theta = 0), (\cos(0)=1). (\lim{b\rightarrow\frac{\pi}{6}^{-}}\int_{0}^{b}\tan(3\theta)d\theta=\lim_{b\rightarrow\frac{\pi}{6}^{-}}\left(-\frac{1}{3}\ln|\cos(3b)|+\frac{1}{3}\ln|\cos(0)|\right)) Since (\lim_{b\rightarrow\frac{\pi}{6}^{-}}\cos(3b)=0) and (\lim_{x\rightarrow0^{+}}\ln x=-\infty), (\lim_{b\rightarrow\frac{\pi}{6}^{-}}\left(-\frac{1}{3}\ln|\cos(3b)|\right)=\infty)

Answer:

B. The integral diverges because (\int_{0}^{\frac{\pi}{6}}\tan(3\theta)d\theta=\infty)