Problem 52 - Calculus Single Variable

Question

Evaluate the given integral. $$ \int_{\pi / 6}^{\pi / 4} \cos (2 x) d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \frac{2-\sqrt{3}}{4} \).

Step 1: Identify the integral formula

The given integral is \( \int \cos(2x) \, dx \). We know from calculus that the integral of \( \cos(ax) \) is \( \frac{1}{a} \sin(ax) + C \), where \( C \) is the constant of integration.

Step 2: Apply the integration rule

Using the formula from Step 1, integrate \( \cos(2x) \). We find that:\[\int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C\]

Step 3: Evaluate the definite integral

Substitute the limits of integration \( \frac{\pi}{6} \) and \( \frac{\pi}{4} \) into the evaluated integral:\[\left[ \frac{1}{2} \sin(2x) \right]_{\pi/6}^{\pi/4} = \frac{1}{2} \sin\left(\frac{\pi}{2}\right) - \frac{1}{2} \sin\left(\frac{\pi}{3}\right)\]

Step 4: Calculate the sine values

Calculate the values of the sine functions:- \( \sin\left(\frac{\pi}{2}\right) = 1 \)- \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)

Step 5: Compute the final result

Plugging the sine values into the expression from Step 3, we have:\[\frac{1}{2} \cdot 1 - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} - \frac{\sqrt{3}}{4}\]This simplifies to:\[\frac{2}{4} - \frac{\sqrt{3}}{4} = \frac{2 - \sqrt{3}}{4}\]

Step 6: Simplified Final Answer

Therefore, the value of the integral \( \int_{\pi / 6}^{\pi / 4} \cos (2 x) \, dx \) evaluates to \( \frac{2 - \sqrt{3}}{4} \).

Key Concepts

Integration by Substitution

Integration by substitution is a method used in calculus to simplify finding antiderivatives. It is akin to reversing the chain rule for differentiation. In the problem at hand, the function involves the cosine of a multiple of the variable, specifically, \( \cos(2x) \). This is a prime example where substitution aids in evaluating an integral more easily.
To employ substitution, you would typically choose a new variable \( u \) to represent a component of the integrand. Here, we set \( u = 2x \), so \( du = 2 \, dx \). We solve for \( dx \) giving us \( dx = \frac{1}{2} \, du \). Substituting these into the integral, we have:

• The integral \( \int \cos(2x) \, dx \) becomes \( \int \cos(u) \cdot \frac{1}{2} \, du \).

Now integrate \( \cos(u) \) with respect to \( u \), remembering to multiply by the constant factor \( \frac{1}{2} \). The integral \( \int \cos(u) \, du \) is \( \sin(u) \), leading to \( \frac{1}{2} \sin(u) \), which translates back to \( \frac{1}{2} \sin(2x) \) when writing in terms of \( x \). The substitution thus helps simplify and solve the integral efficiently in this scenario.

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent frequently appear in calculus problems and integrals. They are crucial for understanding various physical phenomena and solving problems involving periodic behavior. In our example, \( \cos(2x) \) requires us to recall the integration and properties of trigonometric functions.
Cosine is a function that oscillates between -1 and 1. When faced with the integral of cosine, we use the basic integration formula \( \int \cos(ax) \, dx = \frac{1}{a} \sin(ax) + C \). This formula results from the relationship between the differential of sine and cosine, i.e., the derivative of \( \sin \) is \( \cos \), and that of \( \cos \) is \(-\sin \).
When evaluating a definite integral like \( \int_{\pi/6}^{\pi/4} \cos(2x) \, dx \), understanding the periodic behavior helps compute the difference between sines at different angles quickly:

• \( \sin\left(\frac{\pi}{2}\right) = 1 \) because it is the peak value of sine's cycle.
• \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), which is derived from sine's behavior in trigonometric circles or unit circles.

These properties are essential in evaluating and manipulating trigonometric expressions in calculus.

Calculus Techniques

In calculus, understanding and applying appropriate techniques for integration are vital. The integral presented involves definite integration and requires knowledge of evaluating antiderivatives with specific limits.
When working with definite integrals such as \( \int_{\pi/6}^{\pi/4} \cos(2x) \, dx \), we seek to find the exact area under the curve from \( x = \frac{\pi}{6} \) to \( x = \frac{\pi}{4} \). Calculus techniques here primarily refer to effectively managing substitution as shown, and calculating the net area using fundamental theorems:

• The Fundamental Theorem of Calculus links the concept of differentiation and integration, allowing us to evaluate definite integrals using antiderivatives.
• Compute \( F(b) - F(a) \), where \( F(x) \) is the antiderivative of \( f(x) \).

Evaluating \( \left[ \frac{1}{2} \sin(2x) \right]_{\pi/6}^{\pi/4} \) involves substituting the limits into the antiderivative and finding the difference. This whole process underscores a key calculus tenet: transforming the area-under-curve problem into a straightforward calculation using antiderivatives.