Question

evaluate the following integral or state that it diverges.
int_{\frac{3}{pi}}^{infty}\frac{4}{x^{2}}sec^{2}left(\frac{1}{x}
ight)dx
select the correct choice and, if necessary, fill in the answer box to complete your choice.
a. the improper integral converges and int_{\frac{3}{pi}}^{infty}\frac{4}{x^{2}}sec^{2}left(\frac{1}{x}
ight)dx=square . (type an exact answer, using radicals as needed.)
b. the improper integral diverges.

Explanation:

Step1: Use substitution

Let $u = \frac{1}{x}$, then $du=-\frac{1}{x^{2}}dx$. When $x = \frac{3}{\pi}$, $u=\frac{\pi}{3}$; as $x
ightarrow\infty$, $u
ightarrow0$. The integral $\int_{\frac{3}{\pi}}^{\infty}\frac{4}{x^{2}}\sec^{2}(\frac{1}{x})dx$ becomes $- 4\int_{\frac{\pi}{3}}^{0}\sec^{2}(u)du$.

Step2: Evaluate the integral

We know that the antiderivative of $\sec^{2}(u)$ is $\tan(u)$. So, $-4\int_{\frac{\pi}{3}}^{0}\sec^{2}(u)du=-4[\tan(u)]_{\frac{\pi}{3}}^{0}$.

Step3: Apply the fundamental theorem of calculus

$-4(\tan(0)-\tan(\frac{\pi}{3}))=-4(0 - \sqrt{3}) = 4\sqrt{3}$.

Answer:

A. The improper integral converges and $\int_{\frac{3}{\pi}}^{\infty}\frac{4}{x^{2}}\sec^{2}(\frac{1}{x})dx = 4\sqrt{3}$