Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Perform the indicated integrations. $$ \int \frac{y}{\sqrt{16-9 y^{4}}} d y $$

Step-by-Step Solution

Answer

\( \frac{1}{6} \arcsin\left(\frac{3y^2}{4}\right) + C \)

Step 1: Identify the Type of Integral

The given integral is \( \int \frac{y}{\sqrt{16 - 9y^{4}}} \ dy \). It involves a function with \( y \) in both the numerator and under the square root in the denominator. This suggests the use of a trigonometric or hyperbolic substitution to simplify the expression.

Step 2: Choose an Appropriate Substitution

To simplify \( 16 - 9y^4 \), use the substitution \( y^2 = \frac{4}{3} \sin{\theta} \). This choice works well because the square root in the denominator resembles a trigonometric identity form.

Step 3: Calculate the Derivative for Substitution

Differentiate the expression \( y^2 = \frac{4}{3} \sin{\theta} \) to find \( dy \). First, \( 2y \cdot dy = \frac{4}{3} \cos{\theta} \cdot d\theta \). Thus, \( dy = \frac{2}{3y} \cos{\theta} \cdot d\theta \).

Step 4: Substitute into the Integral

Replace \( y^2 \) and \( dy \) in the original integral. The integral becomes \( \int \frac{y}{\sqrt{16 - 9\left(\frac{4}{3}\sin{\theta}\right)^2}} \cdot \frac{2}{3y} \cos{\theta} \ d\theta \). Simplify this to \( \int \frac{2}{3} \cdot \frac{\cos{\theta}}{\sqrt{16 - 9\left(\frac{4}{3}\sin{\theta}\right)^2}} \ d\theta \).

Step 5: Simplify the Denominator

The denominator simplifies as follows: \( 16 - 9 \left(\frac{4}{3}\sin{\theta}\right)^2 = 16 - 16 \sin^2{\theta} \) which becomes \( 16(1 - \sin^2{\theta}) = 16\cos^2{\theta} \).

Step 6: Simplify the Integral

The integral now becomes \( \int \frac{2}{3} \cdot \frac{\cos{\theta}}{4\cos{\theta}} \ d\theta = \int \frac{1}{6} \ d\theta \).

Step 7: Integrate

Integrate \( \int \frac{1}{6} \ d\theta \) to get \( \frac{1}{6}\theta + C \), where \( C \) is the constant of integration.

Step 8: Back-Substitute \(\theta\)

Recall \( y^2 = \frac{4}{3} \sin{\theta} \), \( \sin{\theta} = \frac{3y^2}{4} \). Therefore, \( \theta = \arcsin\left(\frac{3y^2}{4}\right) \). Substitute back to get \( \frac{1}{6}\arcsin\left(\frac{3y^2}{4}\right) + C \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a useful technique in integral calculus for solving integrals involving radicals or quadratic expressions. The key idea is to transform the integral into a simpler form using trigonometric identities. Here, when faced with a square root like \( \sqrt{16 - 9y^4} \), we can employ trigonometric substitution to make the expressions more manageable.
Instead of dealing with \( y \) directly, we substitute \( y^2 = \frac{4}{3} \sin{\theta} \). This approach leverages the identity \( 1 - \sin^2{\theta} = \cos^2{\theta} \).
This substitution simplifies the integral, ultimately leading to the resolution of complex expressions into basic trigonometric forms. The right choice of substitution often aligns with recognizable trigonometric identities, making it easier to simplify and compute the integral later on.

Definite and Indefinite Integrals

In calculus, integrals can be either definite or indefinite. An indefinite integral represents a family of functions \( F(x) \) such that \( F'(x) = f(x) \). It's usually followed by a constant \( C \) indicating any antiderivative. In this exercise, \( \int \frac{y}{\sqrt{16 - 9 y^{4}}} \, dy \) is an indefinite integral.
Once integrated, we get an expression in terms of \( \theta \), with the result \( \frac{1}{6}\theta + C \).
On the other hand, a definite integral includes limits of integration and provides the exact area under the curve. Here, we're focusing on indefinite integrals, which allows us to express the solution in its generalized form, accounting for all possible constants when back-substituting for the original variable. The use of integration constant \( C \) indicates all the shiftings up or down possible for the function.

Calculus Techniques

Multiple calculus techniques are necessary for solving integral problems like the one in this exercise. Here are a few critical strategies used in calculus that help simplify the integral process:

• Substitution: Apart from trigonometric substitution, standard substitution is also a powerful tool. It alters variables enabling easier integration.
• Algebraic Manipulation: Simplifying expressions before integration is often vital. It could involve factoring, completing the square, or utilizing known identities.
• Integration by Parts: Although not used here, it's another important technique involving products of functions.

In the exercise, after the substitution, the integral became simple enough to solve with basic integration commands. Proper application of these calculus techniques effectively simplifies and unlocks the solution to seemingly complex integrals, making the process systematic and approachable.