Integral of Arctan Definitions and Examples

Integral of Arctan Definitions, Formulas, & Examples

GET TUTORING NEAR ME!

(800) 434-2582

By submitting the following form, you agree to Club Z!'s Terms of Use and Privacy Policy

    Integral of Arctan Definitions and Examples

    The integral of arctan, also called the arc tangent function, is a mathematical function that calculates the arc length of a curve from the angle between the tangent line and the x-axis. This function is used in many applications, such as calculating the area of a circular sector or determining the curvature of a curve. In this blog post, we will explore the definition and examples of the integral of arctan. We will also discuss its applications and how it can be used to solve problems.

    Integral of Arctan (Tan Inverse x)

    In mathematics, an integral of the form ? arctan(x) dx is called an inverse trigonometric function. It is a particular case of an improper integral and is defined as follows:

    ? arctan(x) dx = ? 1/(1+x2) dx

    The domain of this function is the set of all real numbers x such that -1 < x < 1. The range of this function is the set of all real numbers y such that -?/2 < y < ?/2.

    This function has a number of properties that make it useful in calculus and other mathematical disciplines. For example, it can be used to find the area under the curve y = 1/(1+x2). It can also be used to solve differential equations.

    If you are unfamiliar with integrals, you may want to review the definition and examples before reading further.

    What is the Integral of Arctan?

    The integral of arctan is defined as the area under the curve of y=arctan(x) from x=-? to x=+?. This area can be found using Calculus by finding the limit as n approaches infinity of the Riemann sum:

    S_n=?_(i=1)^n?((arctan(xi))-(arctan(x_i-1)))?(x_i-x_i-1)

    However, this integral does not have a closed form solution and must be approximated numerically.

    Integral of Arctan Formula

    The integral of arctan can be expressed in terms of the natural logarithm as follows:

    ? arctan(x) dx = ln|sec(x)| + C

    This formula can be derived by first noting that:

    ? arctan(x) dx = ? 1/(1+x^2) dx

    Which can then be rewritten as:

    ? 1/(1+x^2) dx = ? 1 – x^2/(1+x^2) dx
    = ? 1 – (1-1/[1+x^2])*[1/[1+x^2]]dx
    = ? 1 – (1-[arctan(x)]^2)/[1+x^2] dx #By the definition of arctan(x). [arctan(0)=0]

    Integral of Arctan Proof Using Integration by Parts

    The process of integration by parts is a method to find the integral of a function that is the product of two other functions. In this case, we will be using it to find the integral of arctan(x). We will do this by first breaking up arctan(x) into two functions: g(x)=arctan(x) and h(x)=1. Then, we will take the derivative of g(x), which is g'(x)=1/(1+x^2), and multiply it by h(x). Next, we will take the derivative of h(x), which is h'(x)=0, and multiply it by g(x). Finally, we will add these two equations together and solve for the integral.

    So, our final equation will be:

    g'(x)*h(x)+g(x)*h'(x)=Integral of arctan(x)

    Which can be simplified to:

    1/(1+x^2)*1+arctan(x)*0=Integral of arctan (X)

    Definite Integral of Tan Inverse x From 0 to 1

    The integral of tan inverse x from 0 to 1 is defined as the area under the curve of y = tan inverse x from 0 to 1. This area can be approximated using a Riemann sum:

    Let n be the number of rectangles used to approximate the area, and let’s say that each rectangle has a width of w. Then, the height of each rectangle is given by:

    h=tan^{-1}(0)+\frac{1}{n}\times \sum_{i=1}^{n}tan^{-1}\left(\frac{i}{n}\right)

    where i goes from 1 to n. The width w is then given by:

    w=\frac{1}{n}

    So, the total approximation for the integral of tan inverse x from 0 to 1 is given by:

    I_{n}=\sum_{i=1}^{n}h\times w=\frac{1}{n}\times \sum_{i=1}^{n}\left[tan^{-1}(0)+\frac{1}{n}\times \sum_{i=1}^{n}tan^{-1}\left(\frac{i}{n}\right)\right]

    The definition of an integral of arctan

    An integral of arctan is defined as the limit of a Riemann sum where the terms approach zero as the number of terms approaches infinity. This limit can be evaluated using calculus and is equal to the product of the natural logarithm of the tangent function and the constant pi.

    The different types of integrals of arctan

    There are many different types of integrals of arctan, each with its own definition and example.

    The first type of integral is the indefinite integral, which is defined as the antiderivative of a function. The indefinite integral of arctan x is x*arctan(x) + C.

    The second type of integral is the definite integral, which is defined as the limit of a Riemann sum. The definite integral of arctan x from 0 to 1 is 1/2*pi.

    The third type of integral is the improper integral, which is defined as the limit of a certain sequence. The improper integral of arctan x from 0 to infinity is pi/2.

    Examples of integrals of arctan

    -The definite integral of arctan from x=a to x=b is the area under the curve y=arctan(x) between the vertical lines x=a and x=b.
    -The indefinite integral of arctan is any function F(x) such that F'(x)=arctan(x).

    For example,

    ? arctan (x) dx = x – 1/3 x^3 + 1/5 x^5 – …

    Conclusion

    In summary, there are two main types of integrals of arctan – indefinite and definite. Indefinite integrals give the antiderivative (or primitive function) of a function, while definite integrals give the area under a curve between two points. The process for both is similar, although slightly different when it comes to calculating the limits of integration. In general, however, finding the integral of arctan is relatively simple and can be done by using basic algebraic principles.


    Integral of Arctan

    Indefinite integral

    integral tan^(-1)(x) dx = x tan^(-1)(x) - 1/2 log(x^2 + 1) + constant

    Plots of the integral

    Plots of the integral

    Plots of the integral

    Reduced logarithmic form

    1/2 (2 x tan^(-1)(x) - log(1 + x^2)) + constant

    Alternate form of the integral

    -1/2 log(x^2 + 1) + 1/2 i x log(1 - i x) - 1/2 i x log(1 + i x) + constant

    Series expansion of the integral at x = 0

    x^2/2 - x^4/12 + x^6/30 - x^8/56 + O(x^9)
(Taylor series)

    Series expansion of the integral at x = -i

    floor(3/4 - arg(x + i)/(2 π)) (i π - π (x + i) + O((x + i)^4)) + (1/4 (-2 log(1 - i x) + 2 log(x + i) - 4 log(2) - i π) + 1/4 (2 i log(x + i) - 2 i (1 + log(2)) + π) (x + i) + 1/8 (x + i)^2 - 1/48 i (x + i)^3 + O((x + i)^4))

    Series expansion of the integral at x = i

    floor((π - 2 arg(x - i))/(4 π)) (i π + π (x - i) + O((x - i)^4)) + (1/4 (-2 log(i x + 1) + 2 log(x - i) - 4 log(2) + i π) + 1/4 (-2 i log(x - i) + 2 i (1 + log(2)) + π) (x - i) + 1/8 (x - i)^2 + 1/48 i (x - i)^3 + O((x - i)^4))

    Series expansion of the integral at x = ∞

    (π x)/2 + (-log(x) - 1) - 1/(6 x^2) + O((1/x)^3)
(generalized Puiseux series)

    Find the right fit or it’s free.

    We guarantee you’ll find the right tutor, or we’ll cover the first hour of your lesson.