/*! 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 13 Evaluate the integral. $$\int_... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integral. $$\int_{-\pi / 4}^{0} \frac{\sin x}{\cos ^{2} x} d x$$

Short Answer

Expert verified
The value of the definite integral is \(1 - \frac{\sqrt{2}}{2}\).

Step by step solution

01

Change of Variables/Trigonometric Substitution

Observe that the integral involves \(\sin x\ and \cos x\ in its integrand. This suggests a substitution, \(u = \cos x\), that will simplify the integral. The differential of \(u\), \(du\), is \(-\sin x dx\). So \(\sin x dx = -du\). This change of variable changes the integrand to \(\frac{-1}{u^2}\).
02

Compute the Indefinite Integral

Now find the integral of \(\frac{-1}{u^2}\) with respect to \(u\), which results in \(1/u\).
03

Reverse Substitution

Replace \(u\) with \(\cos x\) to reverse the substitution made in Step 1. The integral now becomes \(\frac{1}{\cos x}\).
04

Evaluate the Definite Integral

Evaluate the integral \(\frac{1}{\cos x}\) from \(-\pi / 4\) to 0. This gives \( \cos 0 - \cos (-\pi /4) = 1 - \frac{\sqrt{2}}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus, focusing on finding the area under curves. It's like adding up small parts to find the whole. In the exercise, we're dealing with a definite integral from \(-\pi / 4\) to 0 with the integrand \(\frac{\sin x}{\cos^2 x}\). The process of integration involves working with functions to accumulate quantities. There are various techniques to perform integration, and in this case, a trigonometric substitution is used to simplify the expression. Trigonometric identities and properties help ease the integration process for complex expressions. For instance, here, we substitute \(u = \cos x\), which transforms our integral into a simpler form that can be integrated easily.
Change of Variables
The change of variables, also known as substitution, is a powerful method used in calculus to simplify complex integrals. In our exercise, we apply a trigonometric substitution to change variables, from \(x\) to \(u\). By letting \(u = \cos x\), we simplify our integrand. The differential \(du = -\sin x \, dx\) lets us substitute and transform this integral into a more straightforward expression in terms of \(u\):- Start with the original integrand \(\frac{\sin x}{\cos^2 x}\)- Substitute \(\sin x \, dx = -du\) to modify the integral to \(\frac{-1}{u^2}\)\.This transformation makes the integration manageable and allows us to proceed to calculate the indefinite integral. The change of variables often helps to simplify the components, aligning them in a more straightforward form.
Definite Integrals
Definite integrals extend the concept of indefinite integrals by incorporating limits of integration, which correspond to the bounds of the area we are calculating. Here, we compute \(\int_{-\pi / 4}^{0} \frac{\sin x}{\cos^2 x} \, dx\) by evaluating the difference between the antiderivative at the upper and lower limits. After performing the substitution and integrating, we return to the original variable \(x\), resulting in \(\frac{1}{\cos x}\). We then evaluate from \(-\pi / 4\) to 0:- Substitute the upper limit \(\cos(0) = 1\)- Substitute the lower limit \(\cos(-\pi/4) = \frac{\sqrt{2}}{2}\)- Final computation: \(1 - \frac{\sqrt{2}}{2}\)Definite integrals provide a specific value, representing the total area, or integral, over a given interval.

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Most popular questions from this chapter

The Omega function is used for risk/reward analysis of financial investments. Suppose that \(f(x)\) is a pdf on \((-\infty, \infty)\) and gives the distribution of returns on an investment. (Then \(\int_{a}^{b} f(x) d x\) is the probability that the investment returns between \(\mathrm{\)} a\( and \)\mathrm{\(b} .\) ) Let \(F(x)=\int_{-\infty}^{x} f(t) d t\) be the cumulative distribution function for returns. Then \(\Omega(r)=\frac{\int_{r}^{\infty}[1-F(x)] d x}{\int_{-\infty}^{r} F(x) d x}\) is the Omega function for the in- vestment. \(\begin{array}{lllll}\text { (a) Compute } & \Omega_{1}(r) & \text { for } & \text { the } & \text { exponential } & \text { distribution }\end{array}\) \(f_{1}(x)=2 e^{-2 x}, 0 \leq x<\infty .\) Note that \(\Omega_{1}(r)\) will be undefined \((\infty)\) for \(r \leq 0\) (b) Compute \(\Omega_{2}(r)\) for \(f_{2}(x)=1,0 \leq x \leq 1\) (c) Show that the means of \(f_{1}(x)\) and \(f_{2}(x)\) are the same and that \(\Omega(r)=1\) when \(r\) equals the mean. (d) Even though the means are the same, investments with distributions \(f_{1}(x)\) and \(f_{2}(x)\) are not equivalent. Use the graphs of \(f_{1}(x)\) and \(f_{2}(x)\) to explain why \(f_{1}(x)\) corresponds to a riskier investment than \(f_{2}(x)\) (e) Show that for some value \(c, \Omega_{2}(r) > \Omega_{1}(r)\) for \(r < c\) and \(\Omega_{2}(r)< \Omega_{1}(r)\) for \(r>c .\) In general, the larger \(\Omega(r)\) is, the better the investment is. Explain this in terms of this example.

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