Question

Evaluate the following integrals. $$\int \frac{d x}{1-\tan ^{2} x}$$

Short answer

Question: Evaluate the integral $$\int \frac{d x}{1-\tan ^{2} x}$$ Answer: $$\int \frac{d x}{1-\tan ^{2} x} = \frac{1}{2} \ln|1 - \tan x| + \frac{1}{2} \ln|1 + \tan x| + C$$

Step-by-step solution

Apply Pythagorean identity

Apply the Pythagorean trigonometric identity to rewrite the integrand: $$\int \frac{dx}{1 - \tan^2 x} = \int \frac{dx}{\sec^2 x - \tan^2 x}$$

Perform substitution

Let \(\tan x = u\), then \(\sec^2 x\, dx = du\). Now, substitute u for tan x: $$\int \frac{dx}{\sec^2 x - \tan^2 x} = \int \frac{du}{1 - u^2}$$

Evaluate the integral

The integral now can be solved using partial fraction decomposition, rewrite the integrand as follows: $$\frac{1}{1 - u^2} = \frac{A}{1-u} + \frac{B}{1+u}$$ where A and B are constants to be determined. Multiplying both sides by \((1 - u)(1+ u)\), we have: $$1 = A(1+u) + B(1-u)$$ Expanding and gathering terms, we have: $$1 = (A + B) - (A - B)u$$ Comparing the coefficients of the equal powers of u, we can find values of A and B: $$A + B = 1 \quad \text{ and } \quad -(A - B) = 0$$ Solving this system of equations, we find A = 1/2 and B = 1/2. Thus, the integral can be written as: $$\int \frac{du}{1 - u^2} = \int \left(\frac{1/2}{1 - u} + \frac{1/2}{1 + u}\right) du$$ Now, integrate each term: $$\int \left(\frac{1/2}{1 - u} + \frac{1/2}{1 + u}\right) du = \frac{1}{2} \int \frac{du}{1 - u} + \frac{1}{2} \int \frac{du}{1 + u}$$ $$= \frac{1}{2} \ln|1 - u| + \frac{1}{2} \ln|1 + u| + C$$

Back-substitute

Now, substitute back \(\tan x = u\): $$= \frac{1}{2} \ln|1 - \tan x| + \frac{1}{2} \ln|1 + \tan x| + C$$ This is the final answer for the given integral: $$\int \frac{d x}{1-\tan ^{2} x} = \frac{1}{2} \ln|1 - \tan x| + \frac{1}{2} \ln|1 + \tan x| + C$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify the integration process involving trigonometric functions.
By substituting trigonometric identities or expressions for variables, you can transform a complex integral into one that is easier to handle.
For instance, in the given problem, we use the identity \(1-\tan^2 x = \sec^2 x - \tan^2 x\) to simplify the integrand.
This approach reduces the integral to a form where substitution becomes a feasible option.

• First, recognize a trigonometric identity that can be applied to simplify the integrand.
• Substitute the trigonometric expression to introduce a new variable.
• Solve the resulting integral, often easier to evaluate.

This method is particularly useful when dealing with integrals that contain square roots or quadratic expressions, making them more manageable.

Partial Fraction Decomposition

Partial fraction decomposition is a powerful algebraic technique used to simplify rational expressions and make them easier to integrate.
When faced with integrals of rational functions, breaking down the expression into simpler fractions can often aid in finding the antiderivative.
In the solution process, the integral \( \int \frac{du}{1 - u^2} \) is decomposed into simpler fractions \( \frac{A}{1-u} + \frac{B}{1+u} \).

• Identify the rational function and decompose it into partial fractions.
• Write the expression in terms of simpler fractions with constant numerators.
• Solve for the unknown constants through equation matching.
• Integrate each partial fraction separately.

This method streamlines the process of integrating complex rational expressions and is essential when dealing with polynomial denominators.

Pythagorean Identity

Pythagorean identity is a fundamental identity in trigonometry used to relate the squares of sine, cosine, and tangent functions.
The basic Pythagorean identity states that \(\sin^2 x + \cos^2 x = 1\).
For tangent and secant, one commonly used identity is \(\sec^2 x - \tan^2 x = 1\).

• Recognize trigonometric identities to transform the integral.
• Simplify expressions using known identities.
• Use these transformations to facilitate substitution in integrals.

Using these identities effectively is crucial in solving integrals involving trigonometric functions, as it reduces complex trigonometric expressions to more manageable ones.

Definite and Indefinite Integrals

Integrals in calculus are categorized into definite and indefinite integrals, each with different characteristics and applications.
An indefinite integral is often referred to as an antiderivative, representing a general form of integration without upper or lower limits.
It typically includes a constant \(C\) as part of its expression, symbolizing any constant value can be added to an indefinite integral and still satisfy differentiation.

• Definite integrals calculate the actual area under a curve over a specific interval.
• Indefinite integrals do not have specified limits and include the constant of integration \(C\).
• Each type of integral involves different techniques and interpretations, such as the Fundamental Theorem of Calculus for evaluating definite integrals.

Understanding the distinction between these two types of integrals is key to completing integration problems successfully and understanding their real-world applications.