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$$\text { Evaluate the following integrals.}$$ $$\int \tan ^{3} \theta d \theta$$

Short Answer

Expert verified
Answer: The antiderivative of the function $$\int \tan^3 \theta d \theta$$ is $$\frac{1}{2\cos^2{\theta}} + 2\ln|\cos{\theta}| + \frac{\cos^2{\theta}}{2} + C$$.

Step by step solution

01

Rewrite the function in terms of sine and cosine

Rewrite $$\tan^3{\theta}$$ in terms of sine and cosine: $$\int \tan^3 \theta d \theta = \int \frac{\sin^3 \theta}{\cos^3 \theta} d \theta$$
02

Use substitution

Let $$u = \cos \theta$$. Then, $$-du = \sin \theta d \theta$$, and the integral becomes: $$-\int \frac{(-u^2 +1)^2}{u^3} du$$
03

Expand the expression

Expand the expression inside the integral, $$-\int \frac{1 - 2u^2 + u^4}{u^3} du$$ Now, we can separate the terms and integrate: $$-\int \left(\frac{1}{u^3} - 2\frac{1}{u} + u\right)du$$
04

Evaluate the integral

Integrate each term separately: $$-\left(\int \frac{1}{u^3} du - 2\int \frac{1}{u} du + \int u du\right)$$ Now integrate: $$-\left(-\frac{1}{2u^2} - 2\ln |u| + \frac{u^2}{2}\right) + C$$
05

Substitute back

Substitute back $$u = \cos{\theta}$$: $$\frac{1}{2\cos^2{\theta}} + 2\ln|\cos{\theta}| + \frac{\cos^2{\theta}}{2} + C$$ So, the final answer is: $$\int \tan^3 \theta d \theta = \frac{1}{2\cos^2{\theta}} + 2\ln|\cos{\theta}| + \frac{\cos^2{\theta}}{2} + C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Understanding trigonometric identities is essential when dealing with integrals involving trigonometric functions. A trigonometric identity is an equation involving trigonometric functions that holds true for all values of the involved variables. It helps simplify expressions and equations.

For example, the tangent function \( \tan \theta \) can be expressed in terms of sine \( \sin \theta \) and cosine \( \cos \theta \) as follows:
  • \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
This identity is particularly useful when integrating functions like \( \tan^3 \theta \). Rewriting \( \tan^3 \theta \) in terms of sine and cosine allows us to further manipulate the expression for easier integration. Such rewrites transform trigonometric integration problems into polynomial integration tasks, making them more manageable.

Proper utilization of these identities is a foundational skill in calculus and greatly aids in simplifying tough integrals.
Substitution Method
The substitution method, also known as "u-substitution," is a powerful technique used in calculus to simplify integrals. It involves changing the variable of integration to transform the integral into a simpler form.

In this method, a new variable \( u \) is chosen to replace a function within the integral, and this choice should simplify both the function and its derivative. Once the substitution is made, the integral is evaluated with respect to \( u \), and finally, the original variable is substituted back into the solution.

For example, in the integral \( \int \tan^3 \theta d\theta \), by choosing \( u = \cos \theta \) and \( du = -\sin \theta d\theta \), the integral transforms into:
  • \( -\int \frac{(1 - 2u^2 + u^4)}{u^3} du \)
This simplifies the expression by converting a trigonometric function into a rational polynomial function. Using substitution method efficiently requires practice and intuition to recognize patterns and choose the best substitution.
Integral Calculus
Integral calculus focuses on two main operations: finding the antiderivative of a function and calculating the area under a curve. When dealing with integrals, we often encounter expressions that require us to apply various techniques to simplify and evaluate them.

The example \( \int \tan^3 \theta d\theta \) is initially complex, but after employing the substitution method and trigonometric identities, it becomes a set of simpler polynomials.

Integrating a polynomial function, such as \( -\int \left(\frac{1}{u^3} - 2\frac{1}{u} + u\right) du \), involves finding the antiderivative for each term separately:
  • \( \int \frac{1}{u^3} du \) becomes \( -\frac{1}{2u^2} \)
  • \( \int \frac{1}{u} du \) is \( \ln |u| \)
  • \( \int u du \) becomes \( \frac{u^2}{2} \)
Once these individual integrations are completed, they are compiled into the complete antiderivative expression.

Finally, integrating involves constant attention to detail, particularly when substituting back any variables and ensuring the constant of integration \( +C \) is included. Each step in integral calculus builds upon mathematical principles and techniques, aligning the path to a comprehensive solution.

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