# Integral is typical arctan type

• UrbanXrisis
In summary, when substituting x for tan(u), you change the integral to be \int \frac{4}{1+x^2} \ dx = \int \frac{4}{1 + \tan^2 u} \sec^2 u \ du
UrbanXrisis
$$\int_{0}^{1}\frac{4}{1+x^2}$$

$$\int 4(1+x^2)^{-1}$$

$$u=1+x^2$$

$$du=2x$$

$$\frac{2du}{x}=4dx$$

$$2\int \frac{u^{-1}}{x} dx$$

$$2ln(ux)$$

$$=2ln(x+x^3)|_{0}^{1}$$

$$=2ln(1+1^3)-2ln(0+0^3)$$

I know I did something wrong, any ideas?

Yes,your integral is typical $\arctan$ type...You should know this

$$\int \frac{dx}{1+x^{2}}$$

by heart...

Daniel.

If you make a substitution, you need to arrange it such that the resulting expression does not contain the variable you substituted for.

$$\int 4(1+x^2)^{-1} \ dx,$$

$$u = 1 + x^2 \Longrightarrow du = 2x \ dx \Longrightarrow \frac{du}{2x} = dx$$

you now need to express $dx$ in terms of only $u$, ie. with no $x$'s on the left side. You can do this by noting

$$2x = 2\sqrt{(x^2 + 1) - 1} = 2\sqrt{u - 1}$$

so you get

$$dx = \frac{du}{2\sqrt{u - 1}}$$

$$\int \frac{4}{2u\sqrt{u-1}} \ du = \int \frac{2}{u\sqrt{u-1}} \ du$$

which is not very nice. I would advise seeing if you can find a better strategy to start with.

UrbanXrisis said:
$$\frac{2du}{x}=4dx$$

$$2\int \frac{u^{-1}}{x} dx$$

When you make a substitution, you want to express everything in terms of the new variable before integrating. Try instead the substitution:

$$x=tan(u)$$

I've never really learned about acrtan... how does tan(u) fit into this equation?

If u know the substitution metod and apply it for the given function (tangent) u'll learn.

Daniel.

Subsituting $x = \tan{u} \Longrightarrow dx = \sec^2 u \ du$ simplifies the integral completely. Substituting these in gives

$$\int \frac{4}{1+x^2} \ dx = \int \frac{4}{1 + \tan^2 u} \sec^2 u \ du$$

Then all you need to know is

$$\sec^2 u - \tan^2 u = 1$$

which you can prove on your own, and I'm sure you can finish from there

Whozum,you're wrong.

Daniel.

EDIT:Yoiu saw it & deleted the post...

This is a fairly straightfoward arctangent rule as others have said.

$$\int \frac{4}{1 + x^2}dx = 4 \int\frac{1}{1 + x^2}dx = 4\arctan{x} + C$$

The rule for tangent inverses is $$\int\frac{du}{a^2+u^2} = \frac{1}{a}\arctan\frac{u}{a} + C$$

Last edited by a moderator:

Well, since this was a definite integral I thought I'd let the poster apply the bounds, but yes, you are correct. I'll change them...

## 1. What does "Integral is typical arctan type" mean?

The phrase "integral is typical arctan type" refers to a type of indefinite integral that involves the inverse tangent function (arctan). This type of integral is commonly encountered in calculus and mathematical physics.

## 2. How is the typical arctan type integral represented in mathematical notation?

The typical arctan type integral is represented as ∫ arctan(x) dx, where the variable x is the argument of the arctan function and dx represents the differential of x.

## 3. What is the process for solving a typical arctan type integral?

The process for solving a typical arctan type integral involves using trigonometric identities and substitution to simplify the integral, followed by applying integration techniques such as integration by parts or partial fractions.

## 4. What are some real-world applications of the typical arctan type integral?

The typical arctan type integral has various applications in physics, such as in the calculation of electric fields and magnetic fields in certain systems. It is also used in the analysis of alternating current circuits and in solving differential equations in engineering.

## 5. Are there any other types of integrals similar to the typical arctan type?

Yes, there are other types of integrals that involve inverse trigonometric functions, such as the typical arcsin type integral and the typical arccos type integral. These integrals follow a similar process for solving as the typical arctan type integral.

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