/*! 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 42 Evaluate the integrals. $$\int... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals. $$\int \frac{\sin \theta d \theta}{\sqrt{1+\cos ^{2} \theta}}$$

Short Answer

Expert verified
\(-\sinh^{-1}(\cos \theta) + C\).

Step by step solution

01

Identify Substitution

Observe the integral \( \int \frac{\sin \theta \, d\theta}{\sqrt{1 + \cos^2 \theta}} \) and recognize that \( \cos \theta \) is under the squared root. A useful substitution when dealing with \( \sin \theta \) and \( \cos \theta \) is \( u = \cos \theta \). Thus, compute \( du = -\sin \theta \, d\theta \), or equivalently, \( \sin \theta \, d\theta = -du \).
02

Substitute and Simplify

Use the substitution \( u = \cos \theta \). The integral becomes \( \int \frac{-du}{\sqrt{1 + u^2}} \), which simplifies to \( -\int \frac{du}{\sqrt{1 + u^2}} \).
03

Recognize the Standard Integral

The integral \( \int \frac{du}{\sqrt{1 + u^2}} \) is standard and represents the inverse hyperbolic sine function. Therefore, \( \int \frac{du}{\sqrt{1 + u^2}} = \sinh^{-1}(u) + C \).
04

Integrate and Substitute Back

Use the result from Step 3 to write \( -\int \frac{du}{\sqrt{1 + u^2}} = -\sinh^{-1}(u) + C \), where \( C \) is the constant of integration. Substitute back \( u = \cos \theta \) to get the final result: \(-\sinh^{-1}(\cos \theta) + C \).

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a powerful technique in integral calculus that simplifies the integration of expressions involving radicals. When dealing with trigonometric expressions, you often find terms like \( \sqrt{1 + \cos^2 \theta} \) in the integrals. The idea is to substitute a trigonometric identity or variable to rewrite the integral in a more easily solvable form.

In the case of the integral \( \int \frac{\sin \theta \, d\theta}{\sqrt{1 + \cos^2 \theta}} \), a suitable trigonometric substitution is \( u = \cos \theta \). By doing so, you perform a change of variables, shifting from \( \theta \) to \( u \). This adjustment can transform complex trigonometric integrals into simpler algebraic forms.

Here’s how it works step-by-step:
  • Identify the substitution: Look at the term \( \cos^2 \theta \) under the square root, noticing that substituting \( u = \cos \theta \) is helpful because \( du = -\sin \theta \, d\theta \).
  • Restructure the integral: The substitution results in an integral in terms of \( u \), making it easier to handle, as seen in the integral transform \( -\int \frac{du}{\sqrt{1 + u^2}} \).
Trigonometric substitution is frequently used to tackle integrals that involve expressions resembling the Pythagorean identities and can lead directly to recognizable standard forms.
Inverse Hyperbolic Functions
Inverse hyperbolic functions provide a means to solve integrals that are associated with the hyperbolic trigonometric functions. They relate to the natural logarithm and help in recognizing standard integral forms.

Consider the integral \( \int \frac{du}{\sqrt{1 + u^2}} \), which you encounter when applying trigonometric substitution. This is a standard integral form representing the inverse hyperbolic sine function, specifically \( \sinh^{-1}(u) \). Using inverse hyperbolic functions in calculus helps express solutions for integrals that involve algebraic expressions under a square root in a compact form.

Characteristics of Inverse Hyperbolic Functions
  • They are denoted typically by \( \sinh^{-1}, \cosh^{-1}, \) and \( \tanh^{-1} \), among others.
  • These functions relate to the geometry of hyperbolas, much like the ordinary trigonometric functions relate to circles.
This connection makes them very useful and elegantly satisfying because they can express complex relationships using fairly simple expressions. In our solution, recognizing \( \sinh^{-1}(u) \) simplified the integration process substantially.
Definite and Indefinite Integrals
Understanding definite and indefinite integrals is crucial in calculus. They are fundamental concepts used to compute areas under curves, among other applications.

Indefinite integrals, like \( \int \frac{\sin \theta \, d\theta}{\sqrt{1 + \cos^2 \theta}} \), do not specify limits of integration. Consequently, their result includes a constant of integration, denoted as \( C \), because it represents a family of functions. It is the antiderivative or the general solution to the differential.

Features of Integrals in Calculus
  • **Indefinite Integrals**: They lack numerical bounds and yield a general form with an arbitrary constant \( C \).
  • **Definite Integrals**: These have designated limits and result in a specific numerical value representing the area under the curve between those points.
In our particular problem, after applying a series of steps like substitution, we returned to an indefinite integral which expressed the solution as \( -\sinh^{-1}(\cos \theta) + C \). This highlights how indefinite integrals capture the antiderivative aspect, vital in solving calculus problems involving area and accumulation.

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