Question

Why is it when you integrate 1/(1+u^2) du, you would get tan^-1+C? How did it come to that?

Answer 1

let
u=tantheta
du=...??

Answer 2

it actually came from the fact that d/dx arctan(x) = 1/(x^2+1)

Answer 3

@sourwing Is that just a known rule or something?

Answer 4

I don't know what you meant by known rule, but derivative of arctan(x) is derived using implicit differentiation

Answer 5

and the inverse function theorem

Answer 6

If possible, could someone the whole process of getting the derivative of arctanx?

Answer 7

if y = arctan(x), then x = tan(y), then using implicit differentiation,
1 = sec^2(y) dy/dx
dy/dx = 1/cos^2(x) = 1/(x^2 + 1)

Because d/dx (arctanx) = 1/(x^2 + 1), this is why the integral of 1/(x^2+1) is arctan(x).

Answer 8

\[\sec^2(y)=\tan^2(y)+1=x^2+1\]

Answer 9

also, you need the inverse function theorem in order to make sure \[\frac{dy}{dx}\] exists

Answer 10

yeah, I left a lot of details out. But the goal is i'm trying to explain that why integral of arctan(x) is 1/(x^2+1)

Answer 11

you mean the derivative of arctan(x) is 1/(x^2+1)

Answer 12

that too :DD

Answer 13

\[\sin^2 \theta + \cos ^2 \theta =1 \]
divide both sides by cos^2(theta)
\[\tan^2 \theta + 1 = \sec^2 \theta\]

In the integral, we notice if we pick u=tan(theta) we will get a nice simplification.
Take the derivative of u with respect to theta to get: \[\frac{ du }{ d \theta } = \sec^2 \theta\] and multiply both sides by d theta to plug into your integral:

\[\int\limits_{}^{}\frac{ du }{ u^2+1 } = \int\limits_{}^{}\frac{ \sec^2 \theta d \theta }{ \tan^2\theta+1 }=\int\limits_{}^{} d \theta\]

Answer 14

Thank you everyone for your help.

Answer 15

Of course from here since we know u=tan(theta) then theta must be arctan(u). This is the better method since you can easily integrate other similar things. Suppose you have to integrate:

\[\int\limits_{}^{} \frac{ dx }{ x^2 + 9 }\] then what do you do? Do you take the derivative of stuff by hacking in the dark? No. You choose:

x=3tan(theta) since when it plugs in you can easily see you'll get:

\[9\tan^2 \theta + 9 = 9 \sec^2 \theta\]