Problem 279 - Calculus Volume 1

Question

In the following exercises, use a suitable change of variables to determine the indefinite integral. $$ \int t \sin \left(t^{2}\right) \cos \left(t^{2}\right) d t $$

Step-by-Step Solution

Answer

The indefinite integral is \(-\frac{1}{8} \cos(2t^2) + C\).

Step 1: Identify the substitution variable

To simplify the integration, notice the presence of \( t^2 \) inside both the sine and cosine functions. Let \( u = t^2 \). This substitution will simplify the expression and allow us to integrate more easily.

Step 2: Differentiate the substitution variable

Differentiate \( u = t^2 \) to express \( dt \) in terms of \( du \). This gives \( \frac{du}{dt} = 2t \) or equivalently \( du = 2t \, dt \).

Step 3: Express the integral using the substitution

With \( u = t^2 \) and \( du = 2t \, dt \), solve for \( t \, dt \) to obtain \( t \, dt = \frac{1}{2} \, du \). The integral becomes\[ \int t \sin(t^2) \cos(t^2) \, dt = \int \sin(u) \cos(u) \frac{1}{2} \, du. \]

Step 4: Simplify the integral

Factor out the constant \( \frac{1}{2} \) from the integral to get \[ \frac{1}{2} \int \sin(u) \cos(u) \, du. \]

Step 5: Use an identity to simplify the function

Recall the trigonometric identity \( \sin(2x) = 2\sin(x)\cos(x) \). Therefore, \( \sin(u)\cos(u) = \frac{1}{2} \sin(2u) \). Substituting this into the integral gives\[ \frac{1}{2} \int \frac{1}{2} \sin(2u) \, du. \]

Step 6: Integrate the simplified function

The integral is now \( \frac{1}{4} \int \sin(2u) \, du \). The integral of \( \sin(2u) \) is \( -\frac{1}{2} \cos(2u) + C \), where \( C \) is the constant of integration. Hence, the integral becomes \[ -\frac{1}{8} \cos(2u) + C. \]

Step 7: Substitute back the original variable

Since \( u = t^2 \), substitute back to get the integral in terms of \( t \) as \[ -\frac{1}{8} \cos(2t^2) + C. \]

Key Concepts

Change of Variables

When dealing with integrals, sometimes the function inside the integral may seem too complex at first glance. In such cases, the "Change of Variables" method comes to our rescue. This technique is about transforming the integral into a simpler form by replacing the original variable with a new variable.

• For our example, the original variable is \( t \), and our key observation is its repetitive presence inside other functions like sine and cosine.
• Instead of integrating directly with \( t \), we introduce a new variable, \( u \), where \( u = t^2 \). This is a strategic move aimed at simplifying the whole process.
• Once the substitution is made, you differentiate the new variable \( u \) to relate it back to the original \( dt \).

Change of variables is like finding a shortcut for integration. It transforms a seemingly complex operation into an easier task, making our work simpler and less prone to errors.

Trigonometric Substitution

Trigonometric Substitution is a special technique used to evaluate integrals involving expressions like \( \sin(x) \), \( \cos(x) \), and others. It is especially useful when products of sines and cosines appear. In our case, we leverage a trigonometric identity for help:

• This identity, \( \sin(2x) = 2\sin(x)\cos(x) \), helps us rewrite the product of two trigonometric functions, simplifying the integral.
• With \( u = t^2 \), the product \( \sin(u)\cos(u) \) can be represented as \( \frac{1}{2} \sin(2u) \).
• The substitution converts everything into a friendlier trigonometric expression, paving the way to a much simpler integral.

Using trigonometric identities is like updating a confusing puzzle into easy and manageable pieces. It turns something hard to interpret into something that makes perfect sense.

Indefinite Integrals

Indefinite integrals are a type of integral where we seek to find a function for a given derivative. They are like finding out the original function when only the rate of change is known. In our exercise, the indefinite integral is solved as follows:

• We start with the substitution to simplify the integral. Our ultimate goal is to integrate, which means finding an original function from its derivative.
• Practically, solving the integral resulted in finding \( \int \sin(2u) \, du \), which simplifies to \( -\frac{1}{2} \cos(2u) + C \).
• The constant \( C \) is the constant of integration, reminding us that this integral represents a family of functions.
• Finally, never forget to convert back to the original variable \( t \), yielding the solution in familiar terms: \( -\frac{1}{8} \cos(2t^2) + C \).

Indefinite integrals are key in calculus, representing an important capability to reverse the derivative process and discover original functions.