Question

Find the Laplace transform of the given function. \(f(t)=\int_{0}^{t}(t-\tau)^{2} \cos 2 \tau d \tau\)

Short answer

Question: Determine the Laplace transform of the function \(f(t)=\int_{0}^{t}(t-\tau)^{2} \cos 2 \tau d \tau\). Answer: The Laplace transform of the given function is \(F(s) = \frac{1}{4s^{3} + 64s^{3}}\).

Step-by-step solution

Apply Integration by parts twice

We can apply integration by parts in the following way: Let \(u = (t-\tau)^{2}\) and \(dv = \cos 2\tau d \tau\) Then, \(du = -2(t-\tau)d\tau\) and \(v = \frac{1}{2}\sin 2\tau\) Now, we apply integration by parts: \(\int_{0}^{t}(t-\tau)^{2} \cos 2 \tau d \tau = uv|_{0}^{t} - \int_{0}^{t} v du\) Evaluate and simplify expression, then apply integration by parts once again on the resulting integral: \( = (t-\tau)^{2}(\frac{1}{2}\sin 2\tau)|_{0}^{t} - \int_{0}^{t}\frac{1}{2}\sin 2\tau(-2(t-\tau)d \tau\) Let \(u = (t-\tau)\) and \(dv = \sin 2\tau d\tau\) Then, \(du = -d\tau\) and \(v = -\frac{1}{4}\cos 2\tau\) Continue application of integration by parts: \( = (t-\tau)^{2}(\frac{1}{2}\sin 2\tau)|_{0}^{t} + \int_{0}^{t}(t-\tau)(\frac{1}{4}\cos 2\tau d \tau)\)

Evaluate integral

We just need to evaluate the remaining integral and simplify the expression: \( = (t-\tau)^{2}(\frac{1}{2}\sin 2\tau)|_{0}^{t} + (\frac{1}{4})(t-\tau)^{2}(\frac{1}{2}\cos 2\tau)|_{0}^{t} - (\frac{1}{4})\int_{0}^{t}(t-\tau)(\frac{1}{4}\cos 2\tau d \tau)\) \( = (\frac{1}{2}t^{2}\sin 2t - \frac{1}{8}t^{2}\cos 2t \big) - (\frac{1}{16}) \int_{0}^{t}(t-\tau) \cos 2\tau d\tau\)

Apply Laplace transform

We now apply the Laplace transform to the final expression using the following property: \(\mathcal{L}\{f'(t)\} = sF(s) - f(0)\) \(\mathcal{L}\{f(t)\} = \mathcal{L}\{\frac{1}{2}t^{2}\sin 2t\} - \mathcal{L}\{\frac{1}{8}t^{2}\cos 2t\} - \frac{1}{16}\mathcal{L}\{\int_{0}^{t}(t-\tau) \cos 2\tau d\tau\}\) By using the Laplace Transform property and tables: \(F(s) = \frac{1}{2}\mathcal{L}\{t^{2}\sin 2t\} - \frac{1}{8}\mathcal{L}\{t^{2}\cos 2t\} -\frac{1}{16}\mathcal{L}\{\int_{0}^{t}(t-\tau) \cos 2\tau d\tau\}\) \(F(s) = \frac{1}{2}\frac{2!}{s^{3}(s^{2}+4^2)} - \frac{1}{8}\frac{2!}{s^{3}(s^{2}+4^2)} - \frac{1}{16}sF(s)\)

Simplify and obtain final answer

Finally, simplify the expression to obtain the Laplace transform of f(t): \(F(s) = \frac{1}{2(s^{3}(s^{2}+16))} - \frac{1}{4(s^{3}(s^{2}+16))} - \frac{1}{16}sF(s)\) Combine terms and solve for F(s): \(F(s) = \frac{s^{2}+16}{4s^{3}(s^{2}+16)} - \frac{1}{16}sF(s)\) \(F(s) = \frac{1}{4s^{3} + 64s^{3}}\) Therefore, the Laplace transform of the given function is: \(F(s) = \frac{1}{4s^{3} + 64s^{3}}\)

Key concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus used to tackle integrals where the standard methods may not work. It's particularly handy when dealing with integrals of products of functions. The formula for integration by parts is derived from the product rule for differentiation, and it is given as:

• \( \int u \, dv = uv - \int v \, du \)

In this scenario:

• You assign one part of the function as \( u \), and its corresponding derivative \( du \).
• Another part is marked as \( dv \), with its integral becoming \( v \).

This technique simplifies complex integrals by breaking them down into simpler parts. We notice its usage in the original exercise when the function \( (t-\tau)^2 \cos 2\tau \) is decomposed into two separate integrals using this method.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in mathematics, especially in calculus and its applications. These periodic functions describe waves, oscillations, and many other repeating patterns seen in nature, physics, and engineering.

• The cosine function, \( \cos(x) \), is an even function and describes horizontal repetitions.
• The sine function, \( \sin(x) \), is an odd function and complements the cosine function with vertical oscillations.

In the given exercise, the function \( \cos 2\tau \) appears multiple times, showcasing how trigonometric functions interplay with calculus. The periodic nature of these functions can create unique scenarios when integrated or differentiated, often leading to oscillating integrals requiring specific techniques, like integration by parts.

Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. These equations are ubiquitous in mathematics and science because they describe how things change. They can model anything from population growth to the movement of celestial bodies.

• A common form is: \( \frac{dy}{dt} = f(t, y) \), where \( y \) is an unknown function of \( t \)
• Laplacian transformations often simplify solving differential equations by converting them from the time domain to the s-domain (frequency domain).

In the provided exercise, the Laplace transform is used to convert an integral involving differential behavior into an algebraic expression. This transformation simplifies the handling of differential equations, making it easier to solve complex problems by working in a simpler domain.