Problem 28 - Calculus

Question

Use the method of substitution to find each of the following indefinite integrals. $$ \int \frac{z \cos \left(\sqrt[3]{z^{2}+3}\right)}{\left(\sqrt[3]{z^{2}+3}\right)^{2}} d z $$

Step-by-Step Solution

Answer

\( \frac{3}{2} \sin\left(\sqrt[3]{z^2+3}\right) + C \)

Step 1: Identify the function for substitution

We observe that the integrand contains the term \( \sqrt[3]{z^2 + 3} \). Let's set \( u = \sqrt[3]{z^2+3} \), which implies that \( u^3 = z^2 + 3 \).

Step 2: Differentiate the substitution

Differentiate both sides of \( u^3 = z^2 + 3 \) with respect to \( z \) to find \( dz \) in terms of \( du \). This gives \( 3u^2 du = 2z dz \), so \( z dz = \frac{3}{2} u^2 du \).

Step 3: Rewrite the integral using substitution

Substitute \( u = \sqrt[3]{z^2+3} \) and \( dz = \frac{3}{2u^2} du \). The integral becomes:\[ \int \frac{z \cos(u)}{u^2} \cdot \left(\frac{3}{2u^2} du \right) = \int \frac{3}{2u^2} \cdot z \cos(u) du \]

Step 4: Substitute \( z \) in terms of \( u \)

From \( u^3 = z^2 + 3 \), we find \( z^2 = u^3 - 3 \), and thus \( z = \sqrt{u^3 - 3} \). Now, substitute this back into the integral:\[ \int \frac{3 \cdot \sqrt{u^3 - 3} \cdot \cos(u)}{2u^4} du \]

Step 5: Simplify and integrate

The substitution process can get complicated, but observe simplifying patterns in integration.Instead, let's go back to a simpler view:Using \( z dz = \frac{3}{2} u^2 du \) from step 2, our original integral becomes:\[ \int \cos(u) \cdot \frac{3}{2u^2} du \]This simplifies to \[ \frac{3}{2} \int \cos(u) du \], which integrates to:\[ \frac{3}{2} \sin(u) + C \]

Step 6: Substitute back the original variable

Finally, substitute back \( u = \sqrt[3]{z^2+3} \) to express the answer in terms of \( z \):\[ \frac{3}{2} \sin\left(\sqrt[3]{z^2+3}\right) + C \]

Key Concepts

Indefinite Integral

The concept of an indefinite integral is akin to finding the anti-derivative of a function. When you calculate an indefinite integral, you are essentially trying to find a function whose derivative is the given function. The notation for indefinite integrals is often represented by the integral sign \( \int \) followed by the function and the variable of integration.

• It does not have limits, hence the name 'indefinite.'
• The result includes a “+ C,” known as the constant of integration, because derivatives of constants are zero.
• This constant accounts for any vertical shift in potential anti-derivatives.

Understanding indefinite integrals is crucial in calculus as they provide solutions to differential equations and can be used to describe accumulated quantities, such as areas under curves.

Substitution Method

The substitution method is a favorite technique when tackling integrals that seem complex at first glance. This method is a way to simplify an integral by temporarily changing variables, making the integral easier to solve.

Step-by-Step:

• Identify a part of the integrand that can be replaced by a new variable. Typically, you look for an inner function whose derivative also appears in the integrand.
• Let \( u \) be this identified part, such as \( u = \sqrt[3]{z^2+3} \) from the exercise.
• Differentiate \( u \) with respect to \( z \) to find \( dz \) in terms of \( du \).
• Replace \( z \text{ and } dz \) in the integral with the expressions in terms of \( u \text{ and } du \).

By rewriting the integral in terms of \( u \), it often becomes easier to integrate. After integrating, don't forget to substitute back the original variable to get the solution in its original terms.

Calculus

Calculus is a branch of mathematics that studies how things change. It's divided into differential calculus and integral calculus. While differential calculus concerns itself with the concept of derivatives, how to compute them, and their applications, integral calculus is all about accumulation of quantities, area under a curve, and computing sums of infinitely small factors.

• The derivative and the integral are linked by the Fundamental Theorem of Calculus, which underlines the process of differentiation and integration as inverse processes.
• Differential calculus is used to find rates of change, such as slopes of curves.
• Integral calculus helps to find areas, volumes, displacement, and other accumulative quantities.

A solid understanding of calculus allows one to tackle complex mathematical models in physics, engineering, economics, and beyond, providing tools to analyze and comprehend how quantities vary through space and time.