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Chapter 7: Problem 19

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

Short Answer

Expert verified
Question: Evaluate the integral of the inverse tangent function: \(\int \tan ^{-1} x \, d x\). Answer: The integral of the inverse tangent function is: \(\int \tan ^{-1} x \, d x=(\tan^{-1}x)(x)-\frac{1}{2}\ln |1+x^2|+C\).

Step by step solution

01

Choose u and dv

Let \(u =\tan ^{-1} x\) and \(d v=d x\).
02

Calculate du and v

Now, calculate \(d u\) and \(v\): $$\frac{d u}{d x}=\frac{1}{1+x^{2}}$$ So, \(du=\frac{1}{1+x^{2}}dx\). We know that \(v=\int dv\) and in our case, \(dv=d x\), so we have: $$v=\int d x$$ which results in \(v=x\).
03

Apply the integration by parts formula

Applying the integration by parts formula, we have the following: $$\int \tan ^{-1} x \, d x=u v-\int v \, d u$$
04

Substitute u, v, du into the formula

Now, substitute the values of \(u\), \(v\), and \(d u\) we found earlier into the formula: $$\int \tan^{-1}x \, d x=(\tan^{-1}x)(x)-\int x\frac{1}{1+x^{2}}\,d x$$
05

Evaluate the new integral

Now, we need to evaluate the new integral: $$\int x\frac{1}{1+x^{2}}\,d x$$ To solve this integral, we can use a simple substitution method. Let \(y = 1 + x^2\) and \(dy = 2x \, d x\). We can rewrite the integral as follows: $$\frac{1}{2} \int \frac{dy}{y}$$ Now, the integral is easy to evaluate: $$\frac{1}{2} \int \frac{dy}{y} = \frac{1}{2}\ln |y|+C$$ Substitute back the original expression for y: $$\frac{1}{2}\ln |1+x^2|+C$$
06

Combine the results

Combine the results to find the final answer: $$\int \tan ^{-1} x \, d x=(\tan^{-1}x)(x)-\frac{1}{2}\ln |1+x^2|+C$$

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Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions are the opposite operations of the regular trigonometric functions like sine, cosine, and tangent. They provide angles given the ratio of sides. For the function \(\tan^{-1}(x)\), it returns an angle whose tangent is \ x\.
These functions are vital in calculus, particularly in dealing with non-standard integrals, as they frequently appear along with derivatives.
Recognizing the derivatives of inverse trigonometric functions is important for integration. For example, the derivative of \(\tan^{-1} x\) is \(\frac{1}{1+x^2}\), which is a standard result you can memorize for automatic identification in integrals.
The ability to work with such functions allows you to evaluate integrals that would otherwise be challenging to solve using standard trigonometric functions. Relationships with their derivatives facilitate multiple integration techniques, including integration by parts.
Definite and Indefinite Integrals
Integrals are central in calculus and measure the area under a curve. Indefinite integrals find an antiderivative, represented with the \(+ C\) constant, while definite integrals compute the area using upper and lower limits.
Understanding the difference helps determine when to focus on finding general solutions versus precise areas, particularly in a range of applications from physics to economics.
In the solution, we dealt with an indefinite integral since no limits were provided. You seek a general form that antiderives the function. The notation \(\int f(x) \, dx\) prompts you to find a function \(F(x)\) such that \(F'(x) = f(x)\).
This process can often require techniques like substitution or integration by parts when dealing with more complex functions like inverse trigonometric ones.
Substitution Method
The substitution method assists in simplifying integrals by changing variables, making them easier to solve. You're translating the integral into a form where a simple anti-derivative can be observed.
It's particularly useful in cases with composite functions or where the derivative of a component is present elsewhere in the integrand. In our solution, we used a substitution to manage the expression \(x \, \frac{1}{1+x^2}\), choosing \(y = 1 + x^2\) and finding \(dy = 2x \, dx\).
This allowed us to rewrite the integral in terms of \ y \, significantly simplifying the integration process. The solution showed that \(\frac{1}{2} \int \frac{dy}{y}\) could be solved as \(\frac{1}{2}\ln |y|+C\).
Substitution often reveals elegant solutions to challenging integrals, highlighting the power of transformation in calculus.

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Most popular questions from this chapter

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}\) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}.\)

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt{1+\sqrt{x}}} ; x=\left(u^{2}-1\right)^{2}$$

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x}$$

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int \frac{4 x^{6}}{x^{4}+3 x^{2}} d x\), the first step is to find the partial fraction decomposition of the integrand. b. The easiest way to evaluate \(\int \frac{6 x+1}{3 x^{2}+x} d x\) is with a partial fraction decomposition of the integrand. c. The rational function \(f(x)=\frac{1}{x^{2}-13 x+42}\) has an irreducible quadratic denominator. d. The rational function \(f(x)=\frac{1}{x^{2}-13 x+43}\) has an irreducible quadratic denominator.

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt[4]{x+2}+1} ; x+2=u^{4}$$
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