Problem 33
Question
Evaluate the integrals. $$\int \frac{1}{t^{2}} \cos \left(\frac{1}{t}-1\right) d t$$
Step-by-Step Solution
Verified Answer
Use substitution \( u = \frac{1}{t} \), then evaluate the transformed integral, typically requiring special techniques.
1Step 1: Choose the Appropriate Substitution
We identify a substitution that simplifies the integral. Let \( u = \frac{1}{t} \). Then, \( \frac{du}{dt} = -\frac{1}{t^2} \), and thus, \( dt = -t^2 \, du \). Substitute \( t^2 \) with \( \left(\frac{1}{u}\right)^2 = \frac{1}{u^2}\).
2Step 2: Substitute and Simplify the Integral
Using the substitution outlined in Step 1, replace \( \cos\left(\frac{1}{t}-1\right) \) with \( \cos(u-1) \) and \( dt \) with \( -\frac{1}{u^2} \, du \). The integral becomes:\[\int \cos(u-1) \left(-\frac{1}{u^2}\right) \, du = - \int \frac{\cos(u-1)}{u^2} \, du\]
3Step 3: Evaluate the Simplified Integral
Since the expression involves \( \frac{1}{u^2} \) and trigonometric functions, use integration by parts if necessary or consult a table of integrals for \( \int \frac{\cos(au+b)}{u^2} \, du \). This requires finding an antiderivative or consulting an advanced integration technique.
4Step 4: Write the Final Answer
After evaluating the integral (using methods such as tables, substitution, or integration by parts), re-substitute \( u = \frac{1}{t} \) to express the solution in terms of \( t \).
Key Concepts
Integration by SubstitutionIntegration by PartsTrigonometric Integrals
Integration by Substitution
Integration by substitution is a technique similar to reversing the chain rule of differentiation. It involves replacing a part of the integrand with a new variable to make the integral easier to solve.
To begin, identify a part of the integral that can be substituted with a single variable, usually denoted as \( u \). By choosing \( u = \frac{1}{t} \), the derivative \( \frac{du}{dt} = -\frac{1}{t^2} \) helps in rewriting the entire integral in terms of \( u \).
This method transforms the given complex integral into a new form that's often simpler to integrate. This substitution works well when a function and its derivative appear in the integrand, like in trigonometric integrals. The original exercise shows substitution transforming an integral with \( \cos \left( \frac{1}{t} - 1 \right) \) into a simpler form using \( \cos(u-1) \).
To begin, identify a part of the integral that can be substituted with a single variable, usually denoted as \( u \). By choosing \( u = \frac{1}{t} \), the derivative \( \frac{du}{dt} = -\frac{1}{t^2} \) helps in rewriting the entire integral in terms of \( u \).
This method transforms the given complex integral into a new form that's often simpler to integrate. This substitution works well when a function and its derivative appear in the integrand, like in trigonometric integrals. The original exercise shows substitution transforming an integral with \( \cos \left( \frac{1}{t} - 1 \right) \) into a simpler form using \( \cos(u-1) \).
- Identify which part of the integrand can be transformed to a friendly expression.
- Find \( du \) in terms of \( dt \) or your original variable's differential.
- Rewrite the integral in terms of \( u \).
Integration by Parts
Integration by parts is a powerful method to tackle integrals that consist of a product of two functions. Unlike substitution, which simplifies an integral by changing variables, integration by parts separates an integral into two components.
The formula for integration by parts is derived from the product rule of differentiation and is given by \( \int u \, dv = uv - \int v \, du \). Here, \( u \) and \( dv \) are chosen such that \( du \) and \( v \) are easily determined, simplifying the integration process.
In situations where substitution results in no simplification or additional challenges, as it might with integrals like \( \int \frac{\cos(u-1)}{u^2} \, du \), integration by parts can be the alternative route. This approach often works well when dealing with products involving polynomial and exponential or trigonometric functions.
The formula for integration by parts is derived from the product rule of differentiation and is given by \( \int u \, dv = uv - \int v \, du \). Here, \( u \) and \( dv \) are chosen such that \( du \) and \( v \) are easily determined, simplifying the integration process.
In situations where substitution results in no simplification or additional challenges, as it might with integrals like \( \int \frac{\cos(u-1)}{u^2} \, du \), integration by parts can be the alternative route. This approach often works well when dealing with products involving polynomial and exponential or trigonometric functions.
- Choose \( u \) as a function that simplifies upon differentiation (often a polynomial).
- Let \( dv \) be the rest of the integral, allowing easy integration to find \( v \).
Trigonometric Integrals
Trigonometric integrals involve the integration of functions containing trigonometric functions such as sine, cosine, tangent, and their inverses. These integrals often require specific strategies to solve, like substitution or using trigonometric identities.
In the original exercise, trigonometric functions like \( \cos(u-1) \) appear within the integral. One must look for identities or simplifications that can make the integration process easier. Even a simple identity such as \( \cos(x) = \sin(x+\frac{\pi}{2}) \) could potentially transform the problem.
Sometimes, breaking down the trigonometric functions using identities is the key to solving such integrals or finding antiderivatives in tables. Understanding these nuances helps manage integrals involving functions like \( \sin(ax+b) \), \( \cos(ax+b) \), or their products and quotients.
In the original exercise, trigonometric functions like \( \cos(u-1) \) appear within the integral. One must look for identities or simplifications that can make the integration process easier. Even a simple identity such as \( \cos(x) = \sin(x+\frac{\pi}{2}) \) could potentially transform the problem.
Sometimes, breaking down the trigonometric functions using identities is the key to solving such integrals or finding antiderivatives in tables. Understanding these nuances helps manage integrals involving functions like \( \sin(ax+b) \), \( \cos(ax+b) \), or their products and quotients.
- Be familiar with basic and advanced trigonometric identities.
- Consider substitution if a suitable expression within the trigonometric function can be identified.
- Consult tables of integrals for known antiderivatives of common trigonometric forms.
Other exercises in this chapter
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