Problem 78 - Applied Calculus

Question

Find the indefinite integrals. $$\int\left(2 e^{x}-8 \cos x\right) d x$$

Step-by-Step Solution

Answer

\( \int (2e^x - 8\cos x) \, dx = 2e^x + 8\sin x + C \)

Step 1: Break Down the Integral

The given integral is \( \int (2e^x - 8\cos x) \, dx \). We can rewrite it as two separate integrals: \( \int 2e^x \, dx - \int 8\cos x \, dx \). This will allow us to integrate each term separately.

Step 2: Integrate the Exponential Term

Consider the integral \( \int 2e^x \, dx \). Using the formula for the integral of an exponential function, \( \int e^x \, dx = e^x + C \), the integral becomes \( 2e^x + C \).

Step 3: Integrate the Trigonometric Term

Consider the integral \( \int 8\cos x \, dx \). Using the formula for the integral of a cosine function, \( \int \cos x \, dx = \sin x + C \), the integral becomes \( 8\sin x + C \).

Step 4: Combine the Results

Finally, combine the results of the two integrals from Steps 2 and 3: \( 2e^x - 8\cos x = 2e^x + 8\sin x + C \). Thus, the indefinite integral is: \( \int (2e^x - 8\cos x) \, dx = 2e^x + 8\sin x + C \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique often used to solve integrals involving the product of functions. While not specifically required for the exercise you worked on, knowing this concept can be valuable for more complex integrals.

Here's how it works:

• Choose two functions within the integral: \( u \) (which will be differentiated) and \( dv \) (which will be integrated).
• The formula for integration by parts is: \( \int u \, dv = uv - \int v \, du \).
• Apply the formula carefully by choosing which function to differentiate and which to integrate.

For many problems, the choice of \( u \) and \( dv \) is crucial to simplifying the integration process. Remember, practice makes perfect! Start with simple examples and work your way up to tougher problems. This method is especially useful when dealing with exponential (like \( e^x \)) and trigonometric functions.

Exponential Functions

Exponential functions, especially \( e^x \), are very common in calculus. They have unique properties that make them easier to work with in integration.

• One key feature of the exponential function \( e^x \) is that its derivative and integral are both \( e^x \). This simplicity makes dealing with such functions straightforward compared to others.
• The general form for integrating exponential functions is: \( \int e^{kx} \, dx = \frac{1}{k}e^{kx}+C \). For our task, \( k \) is 1, so \( \int e^x \, dx = e^x + C \).
• Multiplying by a constant, as seen in \( 2e^x \), only requires us to multiply our integral by that constant: \( 2e^x + C \).

This straightforward integration process makes exponential functions one of the less daunting components of calculus. Remember, practice these simple rules and they will soon become second nature.

Trigonometric Functions

Trigonometric functions, such as \( \, cos(x) \, \) and \( \, sin(x) \, \), are prevalent in math problems involving cycles or waves. Integrating such functions requires understanding their basic integral forms.

• The integral of \( \, cos(x) \, \) is \( \, \int \cos x \, dx = \sin x + C \). This is because the derivative of \( \, sin(x) \, \) is \( \, cos(x) \).
• When you have a constant multiplier, such as \( \, 8 \), outside of a trigonometric function, simply multiply the integral by the constant: \( \, \int 8\cos x \, dx = 8\sin x + C \).
• These rules also apply to other trigonometric functions like \( \sin(x), \tan(x) \), and \( \sec(x) \), each with its respective formula.

By mastering these basic integral forms, you simplify many problems involving trigonometric terms. Consider practicing different examples to build confidence and fluency.