Question

In the following integrals express the sines and cosines in exponential form and then integrate to show that: $$\int_{0}^{2 \pi} \sin ^{2} 4 x d x=\pi$$

Short answer

The integral \int_{0}^{2 \pi} \sin^{2} 4x \ dx = \pi \ when evaluated.

Step-by-step solution

Express \( \sin^{2} 4x \) in exponential form

Using the identity \( \sin^{2} \theta = \frac{1 - \cos 2\theta}{2} \), we can express \sin^{2} 4x \ as: \[ \sin^{2} 4x = \frac{1 - \cos 8x}{2} \]

Set up the integral with the new form

Substitute the exponential form from Step 1 into the integral: \[ \int_{0}^{2 \pi} \sin^{2} 4x \ dx = \int_{0}^{2\rm \pi} \frac{1 - \cos 8x}{2} \ dx \]

Separate the integral into two simpler integrals

Rewrite the integrand and separate the integral: \[ \int_{0}^{2 \pi} \sin^{2} 4x \ dx = \frac{1}{2} \int_{0}^{2 \pi} (1 - \cos 8x) \ dx = \frac{1}{2} \left( \int_{0}^{2 \pi} 1 \ dx - \int_{0}^{2 \pi} \cos 8x \ dx \right)\rm \]

Evaluate each integral separately

Compute the integral of the constant \1 \ and \cos 8x\ as follows: \[ \int_{0}^{2 \pi} 1 \ dx = x \bigg|_{0}^{2 \pi} = 2 \pi - 0 = 2 \pi \] \[ \int_{0}^{2 \pi} \cos 8x \ dx = \left. \frac{\s\r\in 8x}{8} \right|_{0}^{2 \pi} = \frac{\rm \sin 16 \pi - \sin 0}{8} = 0 \]

Combine the results

Substitute back the results from Step 4: \[ \int_{0}^{2 \pi} \sin^{2} 4x \ dx = \frac{1}{2} (2 \pi - 0) = \pi \]

Key concepts

exponential form

In mathematics, especially in calculus and trigonometry, expressing functions in exponential form can simplify the integration process. For example, trigonometric identities, such as sine and cosine, can be represented using exponential functions. This often involves Euler's formula, which states: \[ e^{i\theta} = \cos \theta + i\sin\theta \] and \[ e^{-i\theta} = cos \theta - i\sin\theta. \] By combining these, we get: \[ \s\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} \] and \[ \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}. \] Using these identities can transform trigonometric integrals into exponential integrals, which can sometimes be simpler to evaluate.

integration techniques

Integration techniques are critical tools in solving integral calculus problems. One fundamental method is substitution, where a new variable replaces a function of an old variable to simplify the integration. Another common technique is integration by parts, derived from the product rule for differentiation. This method can reduce complex integrals into simpler ones. Partial fraction decomposition is useful for rational functions, breaking them down into simpler fractions. Recognizing when to use these techniques is key to efficiently solving integrals.

trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. These identities are pivotal in simplifying expressions and solving equations. One important identity is the Pythagorean identity: \[ \sin^2\theta + \cos^2\theta = 1. \] Another crucial identity used in integrations is: \[ \sin^2\theta = \frac{1 - \cos 2\theta}{2}. \] This specific identity often simplifies the integration of squares of sine or cosine terms by converting them into a form that is more straightforward to integrate, as seen in the exercise's solution.

definite integrals

Definite integrals represent the area under a curve between two points on the x-axis. They are computed using the fundamental theorem of calculus, which links differentiation and integration. For instance, the definite integral \[ \int_a^b f(x) \ dx \] represents the accumulation of the function’s values from \( a \) to \( b \). For periodic functions like sine and cosine, definite integrals over a full period can often lead to simplifications. In the sample exercise, evaluating \int_0^{2\pi} \cos 8x \ dx \ illustrates this, as it equals zero, simplifying the overall integral evaluation.