Evaluate the following integral using the Fundamental Theorem of Calculus.
Integral from negative 9 pi divided by 2 to 9 pi divided by 2 left parenthesis cosine x minus 5 right parenthesis dx

To evaluate the integral \(\int_{-\frac{9\pi}{2}}^{\frac{9\pi}{2}} (\cos x - 5) \, dx\) using the Fundamental Theorem of Calculus, we proceed as follows:

First, we find the antiderivative of the integrand \(\cos x - 5\).

1. **Antiderivative:**

   The antiderivative of \(\cos x\) is \(\sin x\), and the antiderivative of \(-5\) is \(-5x\).

   Therefore, the antiderivative of \(\cos x - 5\) is:
   \[
   \sin x - 5x + C
   \]
   where \(C\) is the constant of integration, which we do not need for the definite integral.

2. **Apply the Fundamental Theorem of Calculus:**

   According to the Fundamental Theorem of Calculus, if \(F(x)\) is an antiderivative of \(f(x)\), then:
   \[
   \int_{a}^{b} f(x) \, dx = F(b) - F(a)
   \]

   Here, \(f(x) = \cos x - 5\), so \(F(x) = \sin x - 5x\).

   We need to evaluate \(F\) at the upper and lower limits of the integral:
   \[
   F\left( \frac{9\pi}{2} \right) = \sin \left( \frac{9\pi}{2} \right) - 5 \cdot \frac{9\pi}{2}
   \]
   and
   \[
   F\left( -\frac{9\pi}{2} \right) = \sin \left( -\frac{9\pi}{2} \right) - 5 \cdot \left( -\frac{9\pi}{2} \right)
   \]

3. **Evaluate \(F\) at the limits:**

   To find \(\sin \left( \frac{9\pi}{2} \right)\) and \(\sin \left( -\frac{9\pi}{2} \right)\), we use the periodicity of the sine function. Since \(\sin(x)\) has a period of \(2\pi\):
   \[
   \frac{9\pi}{2} = 4\pi + \frac{\pi}{2}
   \]
   So,
   \[
   \sin \left( \frac{9\pi}{2} \right) = \sin \left( \frac{\pi}{2} \right) = 1
   \]

   Similarly,
   \[
   -\frac{9\pi}{2} = -4\pi - \frac{\pi}{2}
   \]
   So,
   \[
   \sin \left( -\frac{9\pi}{2} \right) = \sin \left( -\frac{\pi}{2} \right) = -1
   \]

   Now substitute these into \(F\):
   \[
   F\left( \frac{9\pi}{2} \right) = 1 - 5 \cdot \frac{9\pi}{2} = 1 - \frac{45\pi}{2}
   \]
   \[
   F\left( -\frac{9\pi}{2} \right) = -1 - 5 \cdot \left( -\frac{9\pi}{2} \right) = -1 + \frac{45\pi}{2}
   \]

4. **Compute the definite integral:**
   \[
   \int_{-\frac{9\pi}{2}}^{\frac{9\pi}{2}} (\cos x - 5) \, dx = \left[ F\left( \frac{9\pi}{2} \right) \right] - \left[ F\left( -\frac{9\pi}{2} \right) \right]
   \]
   \[
   = \left( 1 - \frac{45\pi}{2} \right) - \left( -1 + \frac{45\pi}{2} \right)
   \]
   \[
   = 1 - \frac{45\pi}{2} + 1 - \frac{45\pi}{2}
   \]
   \[
   = 2 - \frac{45\pi}{2} - \frac{45\pi}{2}
   \]
   \[
   = 2 - \frac{90\pi}{2}
   \]
   \[
   = 2 - 45\pi
   \]

Thus, the value of the integral is:
\[
\int_{-\frac{9\pi}{2}}^{\frac{9\pi}{2}} (\cos x - 5) \, dx = 2 - 45\pi
\]

Would you like further details or have any questions about this solution?

Here are some related questions you might find interesting:
1. How do you find the antiderivative of a trigonometric function?
2. What is the periodicity of the sine and cosine functions?
3. How does the Fundamental Theorem of Calculus connect differentiation and integration?
4. How do you handle definite integrals with negative limits?
5. What is the significance of the constant of integration in indefinite integrals?
6. Can you explain the properties of sine and cosine functions?
7. How do you apply trigonometric identities to solve integrals?
8. What are the common mistakes to avoid when solving definite integrals?

**Tip:** When evaluating trigonometric integrals, remember the periodic properties of sine and cosine to simplify calculations involving large arguments.

Learn how to evaluate the integral ∫(cos x - 5) dx from -9π/2 to 9π/2 using the Fundamental Theorem of Calculus. This problem involves finding the antiderivative of cos x - 5 and applying the Fundamental Theorem of Calculus for definite integrals. Suitable for advanced undergraduate students and those studying integral calculus.

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