Question

Perform the indicated integrations. $$ \int_{0}^{\pi / 4} \frac{\tan z}{\cos ^{2} z} d z $$

Short answer

The integral evaluates to \(-\frac{1}{2}\).

Step-by-step solution

Identify the Integrand

We have the integral \( \int_{0}^{\pi / 4} \frac{\tan z}{\cos^{2} z} \, dz \). Notice that \( \tan z \) can be expressed as \( \frac{\sin z}{\cos z} \). So the integrand becomes \( \frac{\sin z}{\cos^3 z} \) or equivalently \( \frac{\sin z}{\cos z} \cdot \frac{1}{\cos^2 z} \).

Substitution

Let's use \( u = \cos z \), then \( du = -\sin z \, dz \), which implies \( -du = \sin z \, dz \). When \( z = 0 \), \( u = \cos 0 = 1 \), and when \( z = \frac{\pi}{4} \), \( u = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).

Transform the Integral

Substitute \( u \) into the integral. The new limits are from 1 to \( \frac{1}{\sqrt{2}} \). The integral is transformed to \(-\int_{1}^{1/\sqrt{2}} \frac{1}{u^3} \, du \).

Integrate

To integrate \(-\int_{1}^{1/\sqrt{2}} \frac{1}{u^3} \, du \), we find the antiderivative of \( \frac{1}{u^3} \), which is \(-\frac{1}{2u^2} \). Thus, the integral becomes evaluated as \( \left[ -\frac{1}{2u^2} \right]_1^{1/\sqrt{2}} \).

Evaluate the Definite Integral

Now, compute \( \left[ -\frac{1}{2u^2} \right]_1^{1/\sqrt{2}} \).\(-\frac{1}{2\left(\frac{1}{\sqrt{2}}\right)^2} + \frac{1}{2(1)^2} = -\frac{1}{2 \cdot \frac{1}{2}} + \frac{1}{2} \)This simplifies to: \(-1 + \frac{1}{2} = -\frac{1}{2}\).

Key concepts

Substitution Method

When solving integrals, especially those involving trigonometric functions, the substitution method can simplify the integration process. This technique involves changing variables to transform the integrand into a more manageable form.

• **Purpose**: By substituting a part of the integrand with a single variable, you often simplify the expression, making it easier to integrate.
• **Choosing the substitution**: A good substitution often derives from identifying parts of the integral that resemble derivatives. For the integral \( \int_{0}^{\pi / 4} \frac{\tan z}{\cos^{2} z} \, dz \), we use \( u = \cos z \), because the derivative, \( du = -\sin z \, dz \), matches the integrand's \( \sin z \, dz \) component.
• **Adjusting the limits**: Substituting affects the original integration limits. Calculate the new limits using the substitution. For \( z = 0 \), \( u = \cos 0 = 1 \); for \( z = \frac{\pi}{4} \), \( u = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).

The substitution method is not limited to trigonometric integrals—it is a versatile tool for various integrals, especially those involving composite functions.

Definite Integrals

Definite integrals are a way to calculate the area under a curve within specified bounds. Unlike indefinite integrals, which produce a family of functions as solutions, definite integrals yield a specific numerical value.

• **Understanding the bounds**: The bounds of a definite integral indicate where the integration occurs on the x-axis (or the respective variable axis). In this case, the integral from \(0\) to \(\frac{\pi}{4}\) evaluates the area under the curve \( \frac{\tan z}{\cos^{2} z} \).
• **Integration process**: Once the limits are set, substitute variables if necessary (as we did with \( u = \cos z \)), transform the integrand, and calculate the antiderivative within the new bounds.
• **Evaluation**: After finding the antiderivative, plug in the upper and lower limits into this expression, and subtract the results to find the definite integral's value. This step is crucial for determining the total accumulated quantity, whether it's area, volume, or another measure.

Definite integrals are particularly useful in physics and engineering for calculating net changes, work done, and in various other applications.

Antiderivative Calculation

An antiderivative is essentially the reverse of taking a derivative. It is a function whose derivative yields the integrand. Calculating antiderivatives is fundamental to solving integrals.

Process

To find an antiderivative, apply the basic rules and techniques of integration to the function in the integral.

• **Standard forms**: Recognizing standard derivatives and their corresponding antiderivatives is crucial. For example, the antiderivative of \( \frac{1}{u^3} \) is \( -\frac{1}{2u^2} \).
• **Integration techniques**: Use integration techniques such as substitution or integration by parts to simplify finding antiderivatives. These techniques handle more complex integrands, transforming them into simpler forms.
• **Evaluation and simplification**: Once the antiderivative of the function is found, it is evaluated at the specified limits if it's a definite integral. Simplifying the calculation, as shown in \( \left[ -\frac{1}{2u^2} \right]_1^{1/\sqrt{2}} \), is often required to obtain the final result.

Mastering antiderivative calculation is essential for solving integrals efficiently because it allows one to move seamlessly between derivatives and integrals, understanding the relationship between them.