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$$\text { Evaluate the following integrals.}$$ $$\int \tan ^{2} x d x$$

Short Answer

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Question: Evaluate the integral $$\int \tan^2{x} dx$$. Answer: The integral $$\int \tan^2{x} dx$$ evaluates to $$\tan{x} - x + C$$.

Step by step solution

01

Use Trigonometric Identity

Let's start by rewriting the integrand with the trigonometric identity $$\tan^2{x} = \sec^2{x} - 1$$. We then have: $$\int \tan^2{x} dx = \int (\sec^2{x} - 1) dx$$
02

Break down the integral

Now we separate the terms within the integral: $$\int (\sec^2{x} - 1) dx = \int \sec^2{x} dx - \int 1 dx$$
03

Evaluate individual integrals

Using the known antiderivative rules, we can now evaluate each of the integrals individually: - The antiderivative of $$\sec^2{x}$$ is $$\tan{x}$$ (as the derivative of the tangent function is secant squared). - The antiderivative of $$1$$ is simply $$x$$. Therefore, we have: $$\int \sec^2{x} dx - \int 1 dx = \tan{x} - x$$
04

Include the constant of integration

Finally, we don't forget to include the constant of integration, denoted by $$C$$: $$\tan{x} - x + C$$ So, the final result of the integral is: $$\int \tan^2{x} dx = \tan{x} - x + C$$.

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Key Concepts

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

Trigonometric Integrals
Trigonometric integrals are a fundamental part of calculus, focusing on integrals that involve trigonometric functions such as sine, cosine, tangent, and their reciprocals like secant and cosecant. These types of integrals often require specific strategies and techniques to solve.

A common method for evaluating trigonometric integrals is to use trigonometric identities to simplify the integrands. For example, in the integration of \( \int \tan^2{x} \, dx \), the identity \( \tan^2{x} = \sec^2{x} - 1 \) is used to rewrite the equation in a more manageable form. This step is crucial because it transforms an otherwise difficult integral into one that involves simpler, well-known integrals.

When tackling trigonometric integrals, keep these tips in mind:
  • Familiarize yourself with basic trigonometric identities (e.g., \( \sin^2{x} + \cos^2{x} = 1 \), \( \tan^2{x} + 1 = \sec^2{x} \)).
  • Decompose complex expressions to split them into simpler parts.
  • Look for opportunities to apply u-substitution where necessary to make integration straightforward.
By mastering these strategies, solving trigonometric integrals becomes significantly easier and more intuitive.
Antiderivative
An antiderivative is a function that reverses differentiation, meaning when you derive the antiderivative, you return to the original function. In some texts, it’s also referred to as the indefinite integral.

Knowing how to find an antiderivative is vital in solving integrals. In integral calculus, particularly when evaluating an indefinite integral like \( \int \tan^2{x} \, dx \), finding the antiderivative involves recognizing the basic antiderivative forms. For instance, the antiderivative of \( \sec^2{x} \) is \( \tan{x} \), which is derived from the fact that the derivative of \( \tan{x} \) is \( \sec^2{x} \). Similarly, the antiderivative of \( 1 \) is simply \( x \) because the derivative of \( x \) is \( 1 \).

The constant of integration \( C \) is added to the result of an indefinite integral to account for any constant that was lost in differentiation. Hence, it ensures every possible original function is captured, as constants vanish during differentiation.

Overall, mastering the basic antiderivatives and understanding how they relate to differentiation is crucial for solving integrals efficiently.
Definite and Indefinite Integrals
Integrals in calculus are classified into two main types: definite and indefinite. Understanding the distinction between them is key to solving integrals correctly.An indefinite integral involves finding the antiderivative of a function and is expressed without upper and lower limits, in the form \( \int f(x) \, dx \). It represents a family of functions and includes the constant \( C \). For example, \( \int \tan^2{x} \, dx = \tan{x} - x + C \) is an indefinite integral.

Definite integrals, on the other hand, are evaluated between specified limits \( a \) and \( b \), expressed as \( \int_{a}^{b} f(x) \, dx \). They result in a specific numerical value representing the net area between the function and the x-axis from \( a \) to \( b \). Unlike indefinite integrals, they do not include the constant of integration \( C \) because the limits provide a unique solution.

  • Indefinite integrals provide a general function and include a constant.
  • Definite integrals calculate specific values by evaluating the definite limits.
Recognizing when to apply each type of integral is crucial and comes with practice and familiarity with various functions and integration techniques.

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Most popular questions from this chapter

Let \(a>0\) and let \(R\) be the region bounded by the graph of \(y=e^{-a x}\) and the \(x\) -axis on the interval \([b, \infty).\) a. Find \(A(a, b),\) the area of \(R\) as a function of \(a\) and \(b\) b. Find the relationship \(b=g(a)\) such that \(A(a, b)=2\) c. What is the minimum value of \(b\) (call it \(b^{*}\) ) such that when \(b>b^{*}, A(a, b)=2\) for some value of \(a>0 ?\)

Consider the curve \(y=\ln x\) a. Find the length of the curve from \(x=1\) to \(x=a\) and call it \(L(a) .\) (Hint: The change of variables \(u=\sqrt{x^{2}+1}\) allows evaluation by partial fractions.) b. Graph \(L(a)\) c. As \(a\) increases, \(L(a)\) increases as what power of \(a ?\)

Imagine that today you deposit \(\$ B\) in a savings account that earns interest at a rate of \(p \%\) per year compounded continuously. The goal is to draw an income of \(\$ I\) per year from the account forever. The amount of money that must be deposited is \(B=I \int_{0}^{\infty} e^{-n t} d t,\) where \(r=p / 100 .\) Suppose you find an account that earns \(12 \%\) interest annully and you wish to have an income from the account of \(\$ 5000\) per year. How much must you deposit today?

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d x}{1+e^{x}}$$

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