Problem 13 - University Calculus: Early Transcendentals

Question

Evaluate the indefinite integrals by using the given substitutions to reduce the integrals to standard form. $$\int \sqrt{x} \sin ^{2}\left(x^{3 / 2}-1\right) d x, \quad u=x^{3 / 2}-1$$

Step-by-Step Solution

Answer

Use substitution \( u = x^{3/2} - 1 \) and integrate to get \( \frac{1}{3}(x^{3/2} - 1 - \frac{1}{2}\sin(2(x^{3/2}-1))) + C \).

Step 1: Identify the Given Substitution

Notice that the substitution provided is \( u = x^{3/2} - 1 \). We will use this to simplify the integral.

Step 2: Differentiate the Substitution

Differentiate \( u = x^{3/2} - 1 \) with respect to \( x \). The derivative is \( \frac{du}{dx} = \frac{3}{2}x^{1/2} \). Therefore, \( du = \frac{3}{2}x^{1/2} dx \).

Step 3: Solve for \( dx \)

Rearrange the equation \( du = \frac{3}{2}x^{1/2} dx \) to express \( dx \): \( dx = \frac{2}{3x^{1/2}} du \).

Step 4: Substitute in the Integral

Substitute \( u = x^{3/2} - 1 \) and \( dx = \frac{2}{3x^{1/2}} du \) into the integral \( \int \sqrt{x} \sin^2(x^{3/2}-1) dx \). This becomes \( \int \sqrt{x} \sin^2(u) \cdot \frac{2}{3x^{1/2}} du \).

Step 5: Simplify and Transpose

Notice that \( \sqrt{x} = x^{1/2} \) cancels out upon substituting, reducing the expression to \( \int \frac{2}{3} \sin^2(u) du \).

Step 6: Integrate the Standard Form

We have \( \int \frac{2}{3} \sin^2(u) du \). Use the identity \( \sin^2(u) = \frac{1 - \cos(2u)}{2} \) to rewrite this as \( \frac{2}{3} \int \left( \frac{1 - \cos(2u)}{2} \right) du \).

Step 7: Evaluate the Integral

Now evaluate \( \frac{2}{3} \int \left( \frac{1 - \cos(2u)}{2} \right) du = \frac{2}{3} \left( \frac{1}{2} \int du - \frac{1}{2} \int \cos(2u) du \right) \). This results in \( \frac{2}{3} \left( \frac{1}{2}u - \frac{1}{4} \sin(2u) \right) + C \).

Step 8: Substitute Back For \( u \)

Replace \( u \) with \( x^{3/2} - 1 \) to obtain the final answer: \( \frac{2}{3} \left( \frac{1}{2}(x^{3/2} - 1) - \frac{1}{4} \sin(2(x^{3/2} - 1)) \right) + C \), simplify if needed.

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify integrals, making them easier to evaluate. Imagine you have a complex integral that seems difficult to solve directly. To tackle this, you can cleverly substitute a part of the integral with a new variable. In our example, we substitute using the expression \( u = x^{3/2} - 1 \).
What happens next? By changing variables, you essentially transform the original integral into a potentially simpler one. This involves differentiating your substitution to find \( du \), which is how much \( u \) changes with respect to \( x \).
Remember, it's crucial to rewrite \( dx \) in terms of \( du \) to successfully substitute into the integral. In the given exercise, this step allowed us to switch from the original variable, \( x \), to the new one, \( u \), easing the path to integration.

Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions that are true for all values of the variables for which the functions are defined. In integration, these identities often play a vital role in simplifying expressions, particularly when dealing with trigonometric functions.
A useful identity in this context is \( \sin^2(u) = \frac{1 - \cos(2u)}{2} \), which appears when we have the integral \( \int \sin^2(u) \, du \). This identity helps us express the integral in terms of simpler functions, such as \( 1 \) and \( \cos(2u) \).
Using identities like these is important because they convert integrals that involve trigonometric squares or other complex powers into simpler, more manageable forms. As a result, evaluating the integral becomes straightforward, as you can now integrate basic trigonometric functions or constants.

Integration Techniques

Integration techniques are strategies and methods used to solve integrals. Depending on the form and complexity of an integral, different techniques might be applied to find a solution. Substitution and use of trigonometric identities are just two examples of many techniques.
For instance, while handling \( \int \sin^2(u) \, du \), after using a trigonometric identity, you separate the integral into two components: \( \int \frac{1}{2} \, du \) and \( \int \frac{-1}{2} \cos(2u) \, du \). The first integral directly gives \( \frac{1}{2}u \), whereas the latter involves a straightforward application of the integral of cosine, i.e., \( \int \cos(2u) \, du = \frac{1}{2} \sin(2u) \).
Finally, these integrals can be evaluated separately and then combined for the final result. This method not only makes the task manageable but also highlights the beauty of integral calculus by showcasing how different techniques simplify complex problems.