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Chapter 5: Problem 41

$$ \int x \sec ^{2}\left(x^{2}\right) d x $$

Short Answer

Expert verified
The integral of \( \int x \sec^2(x^2) \, dx \) is \( \frac{1}{2} \tan(x^2) + C \).

Step by step solution

01

Identify the Integration Type

The given integral \( \int x \sec^2(x^2) \, dx \) resembles an integral that could benefit from a substitution method, particularly because the inner function \( x^2 \) has a derivative \( 2x \), which is similar to the multiplicand \( x \) outside.
02

Choose a Suitable Substitution

We choose \( u = x^2 \). Consequently, \( du = 2x \, dx \). This indicates that \( x \, dx = \frac{1}{2}du \). This substitution will simplify the integral significantly.
03

Substitute and Simplify the Integral

Substitute \( x^2 \) with \( u \) and \( x \, dx \) with \( \frac{1}{2}du \) in the integral, transforming it into \[ \int x \sec^2(x^2) \, dx = \int \frac{1}{2} \sec^2(u) \, du. \]
04

Integrate Using Basic Formulas

The integral now is \( \frac{1}{2} \int \sec^2(u) \, du \). We know that the integral of \( \sec^2(u) \) is \( \tan(u) \). Thus, \[ \int \frac{1}{2} \sec^2(u) \, du = \frac{1}{2} \tan(u) + C, \] where \( C \) is the constant of integration.
05

Back-Substitute to the Original Variable

Replace \( u \) with \( x^2 \) to revert back to the original variable, giving \[ \frac{1}{2} \tan(x^2) + C. \] This is the solution to the original integral.

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool in calculus for solving integrals, especially when they involve composite functions. This method involves replacing a complex part of the integral with a single variable to simplify the integration process.

In the exercise provided, we simplified the integral \( \int x \sec^2(x^2) \, dx \) using substitution. By recognizing that \( x^2 \) is a composite function within the integral and its derivative, \( 2x \), resembles another part of the integrand, we set \( u = x^2 \), thus \( du = 2x \, dx \).

Using substitution, we transformed the integral into a simpler form: \( \int \frac{1}{2} \sec^2(u) \, du \). This step reduces the complexity of the integral and makes use of known integration rules. Substitution is extremely helpful because:
  • It reduces complex expressions into simpler ones.
  • It leverages recognition of standard integral forms.
  • It helps to identify substitutions that simplify derivative calculations.
Definite and Indefinite Integrals
In calculus, integrals are classified into two main categories: definite and indefinite integrals. Indefinite integrals are concerned with finding the antiderivative of a function, while definite integrals compute the net area under a curve between specified limits.

In the current exercise, we worked with an indefinite integral, represented as \( \int f(x) \, dx \). The goal of finding an indefinite integral is to determine a general expression that, when differentiated, yields the original integrand. The final result includes a constant of integration, \( C \), since an indefinite integral accounts for all possible antiderivatives.

Understanding the difference between these types of integrals helps in clearly distinguishing when to apply certain properties and techniques:
  • Indefinite integrals result in a family of functions.
  • Definite integrals provide exact values related to areas.
  • Both are foundational in calculus, yet have different applications.
Trigonometric Integrals
Trigonometric integrals involve expressions that contain trigonometric functions such as sine, cosine, and tangent. These types of integrals often require specific strategies like trigonometric identities or substitution to solve.

In our example, the function \( \sec^2(x^2) \) was integrated. Recognizing that \( \sec^2(u) \) integrates directly to \( \tan(u) \) is a crucial step for solving it more efficiently. Each trigonometric function comes with its own integral rules, and familiarity with these makes handling such integrals much simpler.

Here are a few tips for dealing with trigonometric integrals:
  • Know basic trigonometric identities to simplify integrands.
  • Use substitution when the function involves nested trigonometric forms.
  • Recognize standard trigonometric integral forms to expedite solutions.

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Most popular questions from this chapter

(a) Evaluate \(\int\left[x /\left(x^{2}+1\right)\right] d x\). (b) Use a graphing utility to generate some typical integral curves of \(f(x)=x /\left(x^{2}+1\right)\) over the interval \((-5,5)\).

Show that if \(f\) and \(g\) are continuous functions, then $$ \int_{0}^{t} f(t-x) g(x) d x=\int_{0}^{t} f(x) g(t-x) d x $$

(a) Prove that if \(f\) is an odd function, then $$ \int_{-a}^{a} f(x) d x=0 $$ and give a geometric explanation of this result. [Hint: One way to prove that a quantity \(q\) is zero is to show that \(q=-q .]\) (b) Prove that if \(f\) is an even function, then $$ \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x $$ and give a geometric explanation of this result. [Hint: Split the interval of integration from \(-a\) to \(a\) into two parts at \(0 .]\)

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$ V=V_{p} \sin (2 \pi f t) $$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts \((\mathrm{V})\), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). \((1 \mathrm{~Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the \(\boldsymbol{r m s}\) or root-mean-square value of \(V\). By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$ V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}} $$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f\), and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral.] (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{~V}\) at a frequency of \(60 \mathrm{~Hz}\). What is the peak voltage at such an outlet?

Use a trigonometric identity to evaluate the integral. $$ \int \cot ^{2} x d x $$
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