Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-54, perform the indicated integrations. \(\int \frac{y}{\sqrt{16-9 y^{4}}} d y \quad\)

Step-by-Step Solution

Answer

\( \int \frac{y}{\sqrt{16-9y^4}} \, dy = \frac{1}{9} - \frac{y^2}{6} + C \).

Step 1: Identify the type of integration

The integral \( \int \frac{y}{\sqrt{16-9y^{4}}} \, dy \) involves a function under a square root in the denominator. This suggests the use of a trigonometric or hyperbolic substitution for solving the integral.

Step 2: Determine a suitable substitution

Consider using a substitution that simplifies the denominator. Given the form, let \( y^2 = \frac{4}{3} \sin^2(\theta) \). Then, \( y = \sqrt{\frac{4}{3}} \sin(\theta) \) and \( dy = \sqrt{\frac{4}{3}} \cos(\theta) \, d\theta \).

Step 3: Simplify the integral using substitution

Substitute \( y \) and \( dy \) in the integral: \( \int \frac{\sqrt{\frac{4}{3}} \sin(\theta) \cdot \sqrt{\frac{4}{3}} \cos(\theta) \, d\theta}{\sqrt{16 - 9\left(\frac{4}{3}\sin^2(\theta)\right)^2}} \).Simplifying the denominator: \( \sqrt{16 - 16 \sin^4(\theta)} = 4\cos^2(\theta) \).The integral becomes \( \int \frac{4}{9} \sin(\theta) \cos(\theta) \, d\theta \).

Step 4: Integrate the simplified function

Simplify further: \( \int \frac{4}{9} \sin(\theta) \cos(\theta) \, d\theta = \int \frac{2}{9} \sin(2\theta) \, d\theta \).Using the integral of \( \sin(2\theta) \), we have: \( -\frac{1}{9} \cos(2\theta) + C \).

Step 5: Convert back to original variable

To express in terms of \( y \), recall \( \sin(\theta) = \sqrt{\frac{3}{4}} y \) and find \( \theta \) in terms of \( y \).As \( \cos(2\theta) = 1 - 2\sin^2(\theta) \), express the result in terms of \( y \):\( -\frac{1}{9} \left(1 - 2\left(\frac{3}{4} y^2\right)\right) = \frac{1}{9} (1 - \frac{3 y^2}{2}) + C \).

Step 6: Final result

The integral \( \int \frac{y}{\sqrt{16-9y^{4}}} \, dy \) is computed as:\[ \frac{1}{9} - \frac{3y^2}{18} + C = \frac{1}{9} - \frac{y^2}{6} + C \].

Key Concepts

Trigonometric Substitution

Trigonometric substitution is an integration technique used to simplify integrals that contain square roots. This technique is particularly useful when dealing with integrals that resemble the forms of standard trigonometric identities.
Such situations often involve expressions like \(\sqrt{a^2 - x^2}, \sqrt{a^2 + x^2},\) or \(\sqrt{x^2 - a^2}.\)

In this case, we handled the integral \(\int \frac{y}{\sqrt{16-9y^{4}}} dy\), where a trigonometric substitution was efficient due to its denominator resembling \(\sqrt{a^2 - b^2y^4}.\)
The approach selects \(y^2 = \frac{4}{3} \sin^2(\theta)\) as a substitution, because \(\sin^2(\theta)\) offers a convenient square for simplifying under the square root.
Remember to replace both \(y\) and \(dy\) in the original integral, allowing it to morph into a form that is much easier to integrate.

• Use the Pythagorean identities: These help in simplifying expressions.
• Subsitute back to the original variable once solved.

Definite Integrals

Definite integrals find the accumulation of values, or the 'net area', under a curve over a specific interval \( [a, b]\). They result in a numeric value, unlike indefinite integrals which include an arbitrary constant and express a family of functions.
In calculations, one important feature of definite integrals is how boundaries can be evaluated using the Fundamental Theorem of Calculus.

Our example integral, \(\int \frac{y}{\sqrt{16-9y^{4}}} dy\), wasn't solved as a definite integral, but might be approached that way with given boundaries.
When converting an indefinite process to one involving limits, evaluate the antiderivative at each boundary and compute the difference.
Think of it as calculating the exact accumulated change between two points on a function's graph.

• Definite integrals often solve practical problems, like finding areas.
• Calculate using the antiderivative and limits.

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, represent families of functions whose derivative results in the integrand. \(\int f(x) \,dx\) yields an expression plus a constant \(C\).
The constant represents any vertical shift the antiderivative might experience.

In the integration problem, \(\int \frac{y}{\sqrt{16-9y^{4}}} dy\), the final result exemplifies an indefinite integral, shown in \[ \frac{1}{9} - \frac{y^2}{6} + C. \\]
This general solution accounts for all possible solutions by including the arbitrary constant.
The integration process typically employs transformations like trigonometric substitution to simplify, derive formulas and antiderivatives.

• Symbol \(C\) signifies unlimited possible vertical shifts.
• Important in theoretical contexts and for solving differential equations.

Hyperbolic Substitution

Hyperbolic substitution is a less common technique than trigonometric substitution, yet incredibly helpful in particular contexts. It uses hyperbolic functions \(\sinh(x), \cosh(x),\text{ and }\tanh(x)\) which resemble trigonometric functions, but relate to the geometry of hyperbolas instead of circles.
You might use hyperbolic substitution when the integrand includes expressions like \(\sqrt{a^2+y^2}\) or \(\sqrt{y^2-a^2}\), where standard trigonometric substitutions might not simplify effectively.

While not directly applied in solving our integral \(\int \frac{y}{\sqrt{16-9y^{4}}} dy\), recognizing when hyperbolic substitution is advantageous is essential.
This method parallels trigonometric substitution, swapping variables like \(y\) with hyperbolic terms for easier integration.

• Utilize relationships: \(\cosh^2(x) - \sinh^2(x) = 1\).
• Choose hyperbolic substitutions wisely to fit the problem's form.