Help! Anyone know the antiderivative of arctan?

Question

To be honest, I don't know what arctan is exactly, in relation to the other trig functions. Can anyone clue me in? I'm studying for a Cal II test, and the trig is bogging me down!

Thanks so much for your help, everybody! I was really starting to freak out, wondering what I would do if no one answered.

Answer 1

The antiderivative of arctan is the same as the indefinite integral of that function and is x*arctan(x)-(1/2)*ln|1+x^2|.

arctan(x) is an angle; the angle whose tangent is x. In a right triangle with legs h and w, tan(x) = h/w; then x = arctan(h/w).

Answer 2

Arctan is the same thing as saying inverse tangent (tan^-1).
Its anti-derivative is 1/(1+x^2). I dont remember the proof but that is the answer. Hope this helps!

Answer 3

From the reference: integral arctan (ax) dx =
x arctan ax - (log(1+a*2 x^2))/2a

Answer 4

f y = arctan x then x = tan y that is it is inverse of tan x

now we need to integrate arc tanx

let us take arctan x = 1. arctan x

we know d(uv)/dx = udv/dx + v du/dx
integarting both sides(let int be symbol for integrate)
uv = int(udv/dx) + int vdu/dx
let v = x and u = arctan x
dv/dx = 1 du/dx = 1/(1+x^2)

x arctan x = int(arctan x) + int(x/(1+x^2)
so int(arctanx) = x arctanx - int(x/(1+x^2))
= x arctan x - ln(1+x^2)/2
5 seconds ago

Answer 5

arctan is the inverse of tan

Answer 6

arc tan = tan^(-1)


d(tan^(-1) x)/dx = 1/1+x^ 2