Problem 92 - Precalculus

Question

$$\int_{\pi / 4}^{5 \pi / 6} \cos x d x$$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1 - \sqrt{2}}{2} \).

Step 1: Identify the Integral Type

The given exercise asks for the evaluation of a definite integral of the function \( \cos x \) from \( \pi / 4 \) to \( 5\pi / 6 \). This is a straightforward definite integral where we need to find the antiderivative and evaluate it at the upper and lower limits.

Step 2: Determine the Antiderivative

The antiderivative of \( \cos x \) is \( \sin x \). Thus, the antiderivative we'll use for this integral is \( F(x) = \sin x \).

Step 3: Evaluate the Antiderivative at the Upper Limit

Substitute the upper limit \( 5\pi / 6 \) into the antiderivative to find \( F(5\pi / 6) \). The sine of \( 5\pi / 6 \) is \( \sin(5\pi / 6) = \frac{1}{2} \).

Step 4: Evaluate the Antiderivative at the Lower Limit

Substitute the lower limit \( \pi / 4 \) into the antiderivative to find \( F(\pi / 4) \). The sine of \( \pi / 4 \) is \( \sin(\pi / 4) = \frac{\sqrt{2}}{2} \).

Step 5: Compute the Definite Integral

According to the fundamental theorem of calculus, the value of the definite integral \( \int_{\pi/4}^{5\pi/6} \cos x \, dx \) is given by \( F(5\pi / 6) - F(\pi / 4) \). Thus, the result is \( \frac{1}{2} - \frac{\sqrt{2}}{2} \). This simplifies to \( \frac{1 - \sqrt{2}}{2} \).

Key Concepts

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. In other words, if you have a function and you're given its derivative, finding the antiderivative is essentially reversing the derivative process.

For trigonometric functions like cosine, we have some well-known antiderivatives. In the case of the cosine function, the antiderivative is the sine function. This means that if you differentiate sine, you end up with cosine:

• If \( f(x) = \sin x \), then \( f'(x) = \cos x \).

Therefore, when finding the antiderivative of \( \cos x \), we identify it as \( \sin x \). To perform a definite integral, we find the antiderivative first and then use it to determine the area under the curve between the given limits.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, acting as a bridge between these two main concepts of calculus. It consists of two parts, but for definite integrals, we're mostly concerned with the second part.

This part states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then the definite integral of \( f \) from \( a \) to \( b \) is given by:

• \( \int_a^b f(x) \, dx = F(b) - F(a) \)

This theorem allows us to evaluate definite integrals concretely by subtracting the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit. In the exercise, after identifying the antiderivative \( \sin x \), this theorem guided us to calculate \( \frac{1}{2} - \frac{\sqrt{2}}{2} \) by finding \( \sin(5\pi/6) \) and \( \sin(\pi/4) \).

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions such as sine and cosine. These types of integrals frequently appear in calculus because they are foundational to understanding periodic behaviors in mathematics and physics.

For trigonometric integrals, understanding the basic antiderivatives is crucial:

• \( \int \cos x \, dx = \sin x + C \)
• \( \int \sin x \, dx = -\cos x + C \)

In the problem provided, \( \cos x \) is the trigonometric function to integrate. Knowing the antiderivative \( \sin x \) allowed for easy application of the fundamental theorem of calculus. It is these foundational steps—identifying the type of trigonometric function and recalling its antiderivative—that simplify what might initially seem to be a complex integration problem.