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Chapter 7: Problem 39

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

Short Answer

Expert verified
Question: Evaluate the integral $$\int_{0}^{\pi / 2} \sec x \tan x d x$$ or state that it diverges. Answer: The given integral diverges.

Step by step solution

01

Identify the right substitution

To evaluate the integral, we need to pick a substitution that simplifies the expression inside the integral. We notice that the derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\). Therefore, we can choose the substitution: $$u =\sec(x)$$ Now, let's find the derivative of \(u\) with respect to \(x\): $$\frac{du}{dx} = \frac{d(\sec(x))}{dx} = \sec(x)\tan(x)$$
02

Change the limits of integration

Since we are changing the variable from \(x\) to \(u\), we also need to change the limits of integration. The original limits are \(x=0\) and \(x=\pi/2\). Let's find the corresponding limits in terms of \(u\): When \(x=0\): $$u = \sec(0) = 1$$ When \(x=\pi/2\): $$u = \sec(\pi/2) = \infty$$ So, the limits of integration in terms of \(u\) are \(1\) and \(\infty\).
03

Substitute and evaluate

Now, we can rewrite the integral in terms of \(u\). Using the relationship between \(du\) and \(dx\), we get: $$\frac{du}{\sec(x)\tan(x)} = dx$$ Substituting \(u\) and \(dx\) into the integral, we have: $$\int_{1}^{\infty} u d u$$ Now, we can integrate \(u\) with respect to \(u\): $$\frac{1}{2}u^2\Big|_1^\infty = \frac{1}{2} (\infty^2 - 1^2) = \frac{1}{2}(\infty - 1) = \infty$$ Since the integral evaluates to \(\infty\), we conclude that the given integral diverges.

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Most popular questions from this chapter

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$ \begin{array}{l} \text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}= \\ \qquad\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{array} $$

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Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt{x}+\sqrt[3]{x}} ; x=u^{6}$$

Evaluate the following integrals. Consider completing the square. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$

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