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Magazine Chapter 6: Problem 13
Evaluate the given indefinite integral. \(\int e^{2 x} \cos x d x\)
Short Answer
Expert verified
\( \int e^{2x} \cos x \; dx = \frac{1}{5}(e^{2x}\sin x + 2e^{2x}\cos x) + C \).
Step by step solution
01
Choose the Integration Technique
To evaluate \( \int e^{2x} \cos x \; dx \), we can use the integration by parts method. This technique is beneficial for tackling integrals involving the product of two different types of functions, such as an exponential function and a trigonometric function in this case.
02
Identify the Parts for Integration by Parts
In integration by parts, for an integral of the form \( \int u \, dv \), we use \( \int u \, dv = uv - \int v \, du \). Here, let \( u = e^{2x} \) and \( dv = \cos x \, dx \).
03
Differentiate and Integrate Chosen Parts
Differentiate \( u = e^{2x} \) to get \( du = 2e^{2x} \, dx \). Integrate \( dv = \cos x \, dx \) to get \( v = \sin x \).
04
Apply Integration by Parts Formula
Substitute into the integration by parts formula: \[ \int e^{2x} \cos x \, dx = e^{2x} \sin x - \int (\sin x) (2e^{2x}) \, dx \].
05
Simplify the Expression
Simplify the expression: \[ \int e^{2x} \cos x \, dx = e^{2x} \sin x - 2 \int e^{2x} \sin x \, dx \].
06
Use Integration by Parts Again
To evaluate \( \int e^{2x} \sin x \, dx \), rescind it using integration by parts with the same u and dv choices. Let \( u = e^{2x} \), \( dv = \sin x \, dx \), then \( du = 2e^{2x} \, dx \) and \( v = -\cos x \).
07
Apply Integration by Parts to New Integral
Calculating \( \int e^{2x} \sin x \, dx \) gives: \[ \int e^{2x} \sin x \, dx = -e^{2x} \cos x + 2 \int e^{2x} \cos x \, dx \].
08
Substitute Back and Solve Equation
Substitute back \( \int e^{2x} \sin x \, dx \) into our equation from Step 5: \( \int e^{2x} \cos x \, dx = e^{2x} \sin x - 2(-e^{2x} \cos x + 2 \int e^{2x} \cos x \, dx) \).
09
Solve for the Original Integral
Simplify to isolate \( \int e^{2x} \cos x \, dx \): \[ \int e^{2x} \cos x \, dx = \frac{1}{5}(e^{2x} \sin x + 2e^{2x} \cos x) + C \], where \( C \) is the integration constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indefinite Integral
An indefinite integral represents the "anti-derivative" of a function. When you compute an indefinite integral, you are essentially asking, "What function has a derivative that results in the given function?" With indefinite integrals, we include a constant of integration, denoted as \(C\), because the derivative of a constant is zero, making it impossible to determine from the integral itself the original constant in the function without additional information.
- Indefinite integrals provide a family of functions, rather than a specific one.
- They are denoted generally as \( \int f(x) \, dx \).
- In definite integrals, you calculate the anti-derivative and then subtract its values at the boundaries. This is not done with indefinite integrals.
Exponential Function
Exponential functions are characterized by the constant \(e\) raised to a variable power, such as \(e^{2x}\). They are fundamental in mathematics because of their unique properties related to growth and decay processes.
- Exponential functions always involve the constant \(e\), which is approximately 2.718.
- The derivative of the exponential function \(e^{kx}\) is \(ke^{kx}\), meaning it remains proportional to the original function.
- The integral of \(e^{kx}\) follows similarly, resulting in \(\frac{1}{k}e^{kx} + C\).
Trigonometric Function
Trigonometric functions like \(\cos x\) or \(\sin x\) are deeply connected to circles and oscillatory behavior. They are periodic and occur frequently when modeling any kind of wave-like process, from sound waves to the motion of pendulums.
- These functions repeat their values over specific intervals; for cosine and sine, this period is \(2\pi\).
- They can represent the relationship between the angles and lengths in right angled triangles.
- The derivative of \(\cos x\) is \(-\sin x\), and the integral of \(\cos x\) is \(\sin x + C\).
Integration Technique
Integration by parts is a handy technique in calculus used to integrate products of functions, in particular, when each term belongs to different function families, such as an exponential function and a trigonometric function.
- It is based on the product rule for differentiation, albeit in reverse.
- The formula \( \int u \, dv = uv - \int v \, du \) is key, requiring careful choice of \(u\) and \(dv\).
- This method is especially useful when straightforward integration isn't possible.
