Problem 5 - Applied Calculus

Question

Explain how you can tell if substitution can be used to find an antiderivative. $$\int \sin ^{9} t \cos t d t$$

Step-by-Step Solution

Answer

Substitution is possible because \( \sin t \) is the inner function with \( \cos t \, dt \) as its derivative.

Step 1: Identify the Inner Function

In the integral \( \int \sin^{9} t \cos t \, dt \), identify the part of the integrand that could be considered as an inner function. Here, \( \sin t \) is raised to a power and is followed by \( \cos t \, dt \), indicating that \( \sin t \) could be an inner function.

Step 2: Differentiate the Inner Function

Differentiate the potential inner function \( \sin t \) to see if it simplifies another part of the integral. The derivative of \( \sin t \) is \( \cos t \), which matches another part of the integrand, indicating that substitution is possible.

Step 3: Perform Substitution

Let \( u = \sin t \). Then, \( du = \cos t \, dt \). This substitution transforms the integral to a simpler form \( \int u^9 \, du \).

Step 4: Integrate the New Integral

Integrate \( \int u^9 \, du \) using basic power rule techniques. The antiderivative of \( u^9 \) is \( \frac{u^{10}}{10} + C \).

Step 5: Substitute Back

Substitute \( u = \sin t \) back into the antiderivative, resulting in \( \frac{(\sin t)^{10}}{10} + C \). This is the antiderivative of the original integral.

Key Concepts

Antiderivative

Antiderivatives are functions that reverse differentiation. If you have a function and you know its derivative, you can use antiderivatives to find the original function. Thinking of it as running a calculation backward helps. For example, if the derivative of a function is a power of a variable, its antiderivative would reverse this power rule. In our exercise, we sought the antiderivative of \( \sin^9 t \cos t \). The idea of substitution helps us to break down the problem into smaller pieces, making it simpler to apply the antiderivative rules. An easy way to spot if substitution works is to check if the derivative of one part of the function is present elsewhere in the integrand. This match means substitution can effectively simplify the original problem. Once you've identified this, you replace the complex parts with a simple variable, solve the simplified integral, and then switch back to your initial variable. Antiderivatives aim to bring us from differential forms back to the original function form.

Integration Techniques

Integration techniques are the tools used to find antiderivatives. Among these, substitution is powerful because it reduces complex integrals into more manageable forms.

• Identify an 'inner' function within the integral that can transform the original problem.
• Check if the derivative of this 'inner' function matches another part of the integrand.
• Once confirmed, perform substitution by letting \( u \) be the inner function, giving simplicity to the integrand.

The substitution converts the integral into a basic form, usually making it a straightforward power rule integration. After solving the simpler integral, substitute back to meet the original context of the problem. This approach relies heavily on recognizing patterns, which often means spotting a function and its derivative together. The aim is to make complicated integrals simple and solvable using known rules and simplifications.

Trigonometrical Functions

Trigonometric functions like sine and cosine often pair with derivatives to create integrals that look complex initially. Recognizing the relationship between these functions can help choose the right substitution for simplifying integration. Sine and cosine functions are derivatives of each other in a cyclical way: the derivative of \( \sin t \) is \( \cos t \) and vice versa for integration. This property makes substitution easier within trigonometric integrals. In our exercise, identifying \( \sin t \) as an inner function with its derivative \( \cos t \) naturally suggested substitution. These functions appear frequently in calculus because they are periodic and cyclical, important qualities in physical applications and modeling. Understanding their integral and differential properties underscores how calculus can translate into practical problem-solving techniques in real-world phenomena, like waves and oscillations.