Problem 8 - Calculus for Biology and Medicine

Question

Evaluate the indefinite integral by making the given substitution. $$ \int x \cos \left(x^{2}-1\right) d x, \text { with } u=x^{2}-1 $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{2} \sin(x^2 - 1) + C \).

Step 1: Use the given substitution

First, identify that the substitution given is \( u = x^2 - 1 \). Differentiate both sides with respect to \( x \) to find \( du \). This gives us \( du = 2x \, dx \).

Step 2: Solve for \( dx \)

Since our integral \( \int x \cos(x^2-1) \, dx \) involves \( dx \), solve the expression from Step 1 for \( dx \): \[ dx = \frac{du}{2x} \]

Step 3: Substitute and simplify the integral

Now substitute \( u = x^2 - 1 \) and \( dx = \frac{du}{2x} \) into the integral:\[ \int x \cos(u) \left( \frac{du}{2x} \right) = \frac{1}{2} \int \cos(u) \, du \]The \( x \) terms cancel out.

Step 4: Evaluate the integral

The integral \( \frac{1}{2} \int \cos(u) \, du \) is evaluated as:\[ \frac{1}{2} \sin(u) + C \]Remember that the integral of \( \cos(u) \) is \( \sin(u) \).

Step 5: Substitute back to the original variable

Replace \( u \) with \( x^2 - 1 \) to return to the original variable:\[ \frac{1}{2} \sin(x^2 - 1) + C \]

Step 6: Write down the final answer

The indefinite integral of \( \int x \cos(x^2-1) \, dx \) using the substitution method is:\[ \frac{1}{2} \sin(x^2 - 1) + C \]

Key Concepts

Substitution Method

In calculus, the substitution method is a powerful technique used to simplify the process of evaluating integrals. It is particularly useful when the integral involves a function and its derivative, or when the integral can be rewritten in a way that reveals this type of relationship. In the exercise provided, we used the substitution method to transform a complex integral into a more manageable form.

Here's a quick rundown of how the substitution method works:

• Identify the inner function: In our example, this function was identified as \( u = x^2 - 1 \).
• Differentiate to relate \( du \) and \( dx \): This gives us a way to replace \( dx \) in the integral. Differentiating \( u = x^2 - 1 \) with respect to \( x \) provided the expression \( du = 2x \, dx \).
• Solve for \( dx \): Rewriting this gives \( dx = \frac{du}{2x} \), which we substitute back into the integral.
• Transform the original integral: Substitute \( u \) and \( dx \) into the original integral equation to simplify it.

This method can convert an unsolvable integral into a simpler one that can be evaluated using basic integration rules.

Integration Techniques

Integration can be a challenging area of calculus. The good news is that there are several techniques that can be applied to help find solutions. The substitution method used in this exercise is one such technique, often akin to reversing the chain rule.

Some useful integration techniques include:

• Substitution: Often used when the function inside another function dictates the integral's form.
• Integration by Parts: Particularly handy when the integral is a product of two functions, using the formula \( \int u \, dv = uv - \int v \, du \).
• Partial Fractions: Useful for breaking down more complex rational functions into simpler, more manageable pieces.
• Trigonometric Identities: These are employed when dealing with integrals involving trigonometric functions.

Each technique has specific cases where it shines, but a thorough understanding of all of them will empower any student facing calculus problems.

Calculus Problems

Calculus problems, like the one in our exercise, often appear complex at first glance. However, understanding the underlying principles and having a toolkit of techniques makes them much more approachable.

Problems in calculus can be categorized broadly as:

• Differential Calculus: Focuses on rates of change and slopes of curves.
• Integral Calculus: Deals with accumulation of quantities and areas under or between curves.

In the context of our exercise, we worked on an indefinite integral, which is part of integral calculus. By breaking down the functions and recognizing patterns, we were able to simplify and solve the problem using substitution.

Successfully navigating calculus problems involves both mastering the techniques and developing the intuition to know when and how to apply them. With practice, these problems become less intimidating, and more like enjoyable puzzles to solve.