Question

In the following exercises, find each indefinite integral by using appropriate substitutions. $$\int \tan \theta d\theta$$

Short answer

$-\ln |\cos \theta |+C$

Step-by-step solution

Identify the Substitution

To integrate $\int \tan \theta d\theta$, recognize that $\tan \theta =\frac{\sin \theta }{\cos \theta }$. A suitable substitution can be $u=\cos \theta$ since the derivative $\frac{d}{d\theta }(\cos \theta )=-\sin \theta$ is present in the integrand.

Perform the Substitution

Substitute $u=\cos \theta$, thus $du=-\sin \theta d\theta$ or $-du=\sin \theta d\theta$. Rewrite $\int \frac{\sin \theta }{\cos \theta }d\theta$ as $-\int \frac{1}{u}du$.

Integrate the Simplified Expression

The integral $-\int \frac{1}{u}du$ is a standard integral. Compute it as follows: $-\ln |u|+C$, where $C$ is the constant of integration.

Back Substitute to Original Variable

Since $u=\cos \theta$, substitute back to get $-\ln |\cos \theta |+C$.

Key concepts

Integration by Substitution

Integration by Substitution is a powerful technique to simplify integrals, particularly those involving composite functions. This method relies on changing variables to transform a difficult integral into an easier one.
Here's a basic outline of how it works:

• Identify a part of the integrand that can be substituted as a single variable, usually denoted by $u$.
• Compute the differential $du$ from your substitution; this step helps replace the remaining differential part of the integrand.
• Rewrite the original integral in terms of $u$ and $du$, essentially transforming the integral into a simpler form.
• Once the new integral is solved, substitute back to the original variable to express the integral in its original terms.

In our exercise, integrating $\int \tan \theta d\theta$ involves recognizing that $\tan \theta =\frac{\sin \theta }{\cos \theta }$. Substituting $u=\cos \theta$ simplifies the integral by allowing us to use the derivative of cosine, $\frac{d}{d\theta }(-\sin \theta )$, to replace $\sin \theta d\theta$. This turns the problem into a standard integration form, making the process more straightforward.

Trigonometric Functions

Trigonometric Functions play a vital role in calculus, especially integration. These functions describe relationships in triangles and occur frequently in integral problems.
Key trigonometric functions like sine ($\sin \theta$), cosine ($\cos \theta$), and tangent ($\tan \theta$) have specific properties and derivatives that make them suitable for integration techniques. Here’s what you need to know:

• $\sin \theta$ and $\cos \theta$ are the basic continuous periodic functions that form the foundation of trigonometry.
• The derivative of $\sin \theta$ is $\cos \theta$, and the derivative of $\cos \theta$ is $-\sin \theta$, making them particularly useful in substitution.
• Tangent, expressed as $\tan \theta =\frac{\sin \theta }{\cos \theta }$, often leads to more complex integrals that benefit from substitution.

In our problem, knowing that the tangent function can be expressed as a ratio of sine and cosine allowed us to utilize the chain rule effectively during substitution, facilitating the integration process.

Definite vs. Indefinite Integrals

Definite and Indefinite Integrals serve different purposes in calculus. Both involve finding the integral of a function, but each has a unique outcome and use case.

An **indefinite integral** represents a family of functions and includes a constant of integration, $C$. This integral is of the form $\int f(x)dx=F(x)+C$. It describes the antiderivative of a function, providing a general solution without specific limits.

• Useful for finding function expressions.
• The constant $C$ reflects the general nature by allowing for vertical shifts of the function graph.

A **definite integral**, however, computes the accumulation of a quantity between two limits $a$ and $b$. Represented as ${\int }_a^{b}f(x)dx$, it results in a numerical value.

• Used to calculate areas under curves.
• No constant $C$ is involved; the result is a specific value.

In the given exercise, we focus on an indefinite integral, $\int \tan \theta d\theta$, which requires understanding the antiderivative and includes the constant $C$ at the end. This general form is crucial when looking for function expressions over a range of inputs.