/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 36

Evaluate the following integrals or state that they diverge. $$\int_{0}^{\pi / 2} \tan \theta d \theta$$

Short Answer

Expert verified
Answer: The integral diverges.

Step by step solution

01

Determine if the integral is improper or not

We are given the integral: $$\int_{0}^{\pi / 2} \tan \theta d \theta$$ Let us check the value of the integrand at the endpoint of the interval: $$\tan \left(\frac{\pi}{2}\right) = \lim_{\theta \to \frac{\pi}{2}^-} \frac{\sin \theta}{\cos \theta} = \lim_{\theta \to \frac{\pi}{2}^-} \frac{1}{\cos \theta}$$ As \(\theta\) approaches \(\pi/2\) from the left, \(\cos \theta\) approaches 0, so the expression approaches infinity. This means that the integral is an improper integral.
02

Rewrite the integral as a limit

To evaluate an improper integral of the form: $$\int_{a}^{b} f(x) dx$$ where \(f(x)\) is unbounded at \(b\), rewrite it as: $$\lim_{t \to b^{-}} \int_{a}^{t} f(x) dx$$ Applying this to our integral yields: $$\lim_{t \to \frac{\pi}{2}^-} \int_{0}^{t} \tan \theta d \theta$$
03

Find the antiderivative of the integrand

The antiderivative of \(\tan \theta\) is \(\ln |\sec \theta| + C\), where \(C\) is the constant of integration. We can now apply the Fundamental Theorem of Calculus to find that: $$\lim_{t \to \frac{\pi}{2}^-} \left[\ln |\sec \theta|\right]_{0}^{t} = \lim_{t \to \frac{\pi}{2}^-}\left(\ln |\sec t| - \ln |\sec 0|\right)$$
04

Evaluate the limit

Now, let us evaluate the limit: $$\lim_{t \to \frac{\pi}{2}^-}\left(\ln |\sec t| - \ln |\sec 0|\right) = \lim_{t \to \frac{\pi}{2}^-} \ln |\sec t| - \ln |\sec 0|$$ Recall that \(\sec u = \frac{1}{\cos u}\), so $$\lim_{t \to \frac{\pi}{2}^-} \ln |\sec t| = \lim_{t \to \frac{\pi}{2}^-} \ln \frac{1}{\cos t}$$ As \(t\) approaches \(\pi/2\) from the left, \(\frac{1}{\cos t}\) approaches infinity and therefore, \(\ln \frac{1}{\cos t}\) approaches infinity. Since the limit approaches infinity, we can conclude that the integral diverges.

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