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Finding an Indefinite Integral Involving Secant and Tangent In Exercises \(19-32,\) find the indefinite integral. $$ \int \sec ^{3} \pi x d x $$

Short Answer

Expert verified
\(\frac{1}{2\pi}(\tan(\pi x) \sec(\pi x) + \ln|\sec(\pi x) + \tan(\pi x)|) + C\).

Step by step solution

01

Recognition and rearrangement

Rewrite the given integral \(\int \sec^3(\pi x) dx\) as \(\int \sec(\pi x)\sec^2 (\pi x) dx\). This helps to distinguish and identify an integral by parts problem. We will recognize that \(\sec^2(\pi x)\) is the derivative of \(\tan(\pi x)\).
02

Assigning u and dv

In the formula for integration by parts \(\int u dv = uv − \int v du\), we let \(u = sec(\pi x)\) and \(dv = \sec^2(\pi x) dx\). Then to use the formula, we need to compute \(du = \pi \sec(\pi x) \tan(\pi x) dx\) by differentiating \(u\), and \(v = \tan (\pi x)\) by integrating \(dv\).
03

Apply formula for integration by parts

Substitute \(u\), \(v\), \(du\), and \(dv\) into the formula which then becomes \(\int \sec^3(\pi x) dx = \tan(\pi x) \sec(\pi x) − \int \pi \tan^2(\pi x) \sec(\pi x) dx\).
04

Substitute identity for tan^2

Looking at the integral obtained in the right part of the above equation, \(\int \pi \tan^2(\pi x) \sec(\pi x) dx\), replace \(\tan^2(\pi x)\) by \(\sec^2(\pi x) − 1\), which is the Pythagorean identity for secant and tangent. Then we have \(\int \sec^3(\pi x) dx = \tan(\pi x) \sec(\pi x) − \pi \int (\sec^3(\pi x) dx− \sec(\pi x) dx)\).
05

Simplify expression

The equation contains the original integral \(\int \sec^3(\pi x) dx\) on both sides. Shift the \(\pi \int \sec^3(\pi x) dx\) term from the right side to the left side, giving \(2 \pi \int \sec^3(\pi x) dx = \tan(\pi x) \sec(\pi x) + \pi \int \sec(\pi x) dx\). Divide both sides by \(2\pi\) to solve for the original integral, \(\int \sec^3(\pi x) dx\).
06

Integrate ∫ secxdx and Finalize the solution

The remaining integral to compute is \(\int \sec(\pi x) dx\) which is a standard integral with solution \(1/\pi \ln|\sec(\pi x) + \tan(\pi x)| + C\). Substitute this into the expression obtained in Step 5 to obtain the final solution.

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Key Concepts

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

Integration by Parts
Integration by Parts is a powerful technique used to integrate products of functions when the standard integration methods do not apply. Imagine trying to solve a math puzzle where you need to break it down into smaller, more manageable pieces. Similarly, the formula for Integration by Parts, \( \int u dv = uv − \int v du \), essentially breaks down the original integral into two parts - one that is easier to integrate, and another that is reduced to a simpler form through differentiation.

In our exercise \( \int \sec^3(\pi x) dx \), we choose \( u = \sec(\pi x) \) and \( dv = \sec^2(\pi x) dx \) as parts to apply this method intelligently. This strategic choice comes from recognizing that \( \sec^2(\pi x) \) is the derivative of \( \tan(\pi x) \) and therefore the integration process is simplified. The aim of this technique is to continually simplify the integral until it becomes something we can solve directly.
Trigonometric Integrals
Trigonometric Integrals involve the integration of trigonometric functions. These functions, circular functions of angles, are the backbone of solving various calculus problems, especially those involving periodicity and oscillation.

To grasp the indefinite integral \( \int \sec^3(\pi x) dx \), one must understand how trigonometric integrals are solved by converting complex expressions into simpler trigonometric forms. This involves manipulating the integrand with trigonometric identities or using substitution to transform the integral into a more recognizable form.

Recognizing Patterns

In the case of our exercise, recognizing that we can express \( \sec^3(\pi x) \) as \( \sec(\pi x) \sec^2(\pi x) \) reveals a pattern that is key to solving the integral. This pattern recognition is instrumental in applying further integration techniques.
Pythagorean Identities
Pythagorean Identities are essential tools for manipulating and simplifying trigonometric expressions. Originating from the Pythagorean theorem associated with right-angled triangles, these identities provide relationships between the sine, cosine, secant, and tangent functions.

In handling integrals such as \( \int \sec^3(\pi x) dx \) as presented in our exercise, Pythagorean identities allow us to convert \( \tan^2(\pi x) \) into an expression involving \( \sec(\pi x) \)—namely \( \sec^2(\pi x) - 1 \). By replacing \( \tan^2(\pi x) \) with this identity, we essentially reduce the complexity of the integral and make it solvable by standard methods.

Application in Integration

Thus, understanding and applying Pythagorean Identities is critical to tackling trigonometric integrals and often leads to a breakthrough in the simplification process.
Integration Techniques
Integration Techniques encompass a variety of strategies used to calculate antiderivatives and integrals which may not be immediately solvable. For students to navigate complex integrals, they must be adept with multiple methods such as substitution, partial fraction decomposition, trigonometric identities, and Integration by Parts, to name a few.

Our indefinite integral example, \( \int \sec^3(\pi x) dx \), is a prime example of the integration challenges students often face. It demonstrates the need to combine different techniques; starting off with Integration by Parts, then using a Pythagorean Identity for simplification, and finally performing standard integration of \( \sec(\pi x) \) to finalize the solution.

Choosing the Right Tool

It's like having a toolbox at your disposal. The real skill lies not only in knowing the tools but also in selecting the right one for the job at hand. This exercise tests the student's ability to discern which techniques to apply and when, to arrive at the most efficient solution.

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Most popular questions from this chapter

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t .\) The Laplace Transform of \(f(t)\) is defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ when the improper integral exists. Laplace Transforms are used to solve differential equations. In Exercises \(95-102,\) find the Laplace Transform of the function. $$ f(t)=t $$

Extended Mean Value Theorem In Exercises \(91-94\) , apply the Extended Mean Value Theorem to the functions \(f\) and \(g\) on the given interval. Find all values \(c\) in the interval \((a, b)\) such that $$\frac{f^{\prime}(c)}{g^{\prime}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$$ $$ f(x)=\frac{1}{x}, \quad g(x)=x^{2}-4 \quad[1,2] $$

Evaluating a Limit Let \(f(x)=x+x \sin x\) and \(g(x)=x^{2}-4 .\) (a) Show that \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=0\) (b) Show that \(\lim _{x \rightarrow \infty} f(x)=\infty\) and \(\lim _{x \rightarrow \infty} g(x)=\infty\) (c) Evaluate the limit $$\quad \lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}$$ What do you notice? (d) Do your answers to parts (a) through (c) contradict L'Hopital's Rule? Explain your reasoning.

U-Substitution In Exercises 109 and 110 , rewrite the improper integral as a proper integral using the given u-substitution. Then use the Trapezoidal Rule with \(n=5\) to approximate the integral. $$ \int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x, \quad u=\sqrt{x} $$

Arc Length Sketch the graph of the hypocycloid of four cusps \(x^{2 / 3}+y^{2 / 3}=4\) and find its perimeter.

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