Why Is the Antiderivative of 1/(1+x^2) Equal to arctan(x)?

[luitzen]

Integration of 1/(1+x^2)


The antiderivative of 1/(1+x^2) is arctan(x), according to Maple, but why is that? I don't have a clue in which direction to go. Is there a way you can calculate it or is it something you should remember like the antiderivative of 1/x?

 

[HallsofIvy]

Well, what do you know? The derivative of sin(x) is cos(x), the derivative of cos(x) is -sin(x) so, if you do not know that the derivative of tan(x) is $sec^2(x)$, you could derive it from the quotient rule.

That is if y= arctan(x), then x= tan(y) so $dx/dy= sec^2(y)$. But then
$dy/dx= 1/sec^2(y)= 1/sec^2(arctan(y))$
and you could use trig identities to change sec(arctan(y)) to tan(arctan(y))= y.

Or do this: if x= tan(y) then you can think of this as a right triangle with angle y, opposite side x and near side 1. Find the length of the hypotenuse using the Pythagorean theorem, then use "secant = hypotenuse over near side" to find sec(y).

 

[Kreizhn]

Alternatively, this is a problem in "trigonometric substitution," which essentially plays on the Pythagorean theorem. We know that

$$\sin^2(x) + \cos^2(x) = 1$$

right? Also, if one divides by [itex]\sin^2(x) [/tex] or [itex]\cos^2(x) [/tex] we also get the identities<br /> <br /> $$1 + \cot^2(x) = \csc^2(x), \qquad \tan^2(x) + 1 = \sec^2(x)$$<br /> <br /> The trick to trigonometric substitution is to realize that you can use these identities whenever your denominator is of the form<br /> <br /> $1 - x^2, x^2-1, 1+ x^2$<br /> <br /> by letting x be some appropriate trig function. Give it a try and see what happens!<br /> <br /> Edit: Though technically, in the $x^2 -1$ case it would be more prudent to use hyperbolic trig, you can indeed still use normal trig sub.[/itex][/itex]

 

[luitzen]

Thank you for your answer.

I do know the derivatives of sin(x) and cos(x) and I know that than tan(x)=cos(x)/sin(x), and I know how to find the derivative of that. I didn't know about sec(x), but according to Wikipedia it is 1/cos(x). Is that correct?

The problem, however, is to find the antiderivative of 1/(1+x^2), which happens to be arctan of x and your explanation seems to do the reverse.

 

[luitzen]

The trick to trigonometric substitution is to realize that you can use these identities whenever your denominator is of the form

$1 - x^2, x^2-1, 1+ x^2$

This is very useful, thank you.

 

[Mark44]

Thank you for your answer.

I do know the derivatives of sin(x) and cos(x) and I know that than tan(x)=cos(x)/sin(x), and I know how to find the derivative of that. I didn't know about sec(x), but according to Wikipedia it is 1/cos(x). Is that correct?

Yes. You should know the basic trig identities that express all six trig functions in terms of either sine or cosine.

The problem, however, is to find the antiderivative of 1/(1+x^2), which happens to be arctan of x and your explanation seems to do the reverse.

If you are referring to HallsOfIvy's explanation, he is using the idea that antidifferentiation and differentiation are essentially inverse operations.

$$\int \frac{1}{1 + x^2}dx = arctan(x) + C$$
$$\Rightarrow \frac{d(arctan(x))}{dx} = \frac{1}{1 + x^2}$$