Question

How do you evaluate the integral `int 1/(1+x^2)dx` from 0 to `oo`?

Answer 1

`pi/2`.

Explanation:

We know that `int1/(1+x^2)dx=arc tanx+C.`

Then, by the Fundamental Principle of Definite Integral,

`:. I= int_0^oo 1/(1+x^2)dx=lim_(yrarroo) int_0^y 1/(1+x^2)dx`

`=lim_(yrarroo)[arc tanx]_0^y`

`=lim_(yrarroo) [arc tany-arc tan0]`

Since, the `arc tan` fun. is continuous on `RR`, we have,

`I=arc tan{lim_(yrarroo) y}-0`

`=pi/2`.

Answer 2

I got `pi/2`.

Interesting enough, if you realize that `1/(1+x^2)` is even, you could further say that `int_(-oo)^(oo) 1/(1+x^2)dx = pi`.

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Let us first prove that `d/(dx)[arctanx] = 1/(1+x^2)`.

`arctanx = y`

`x = tany`

Now, we perform implicit differentiation to get

`1 = sec^2y((dy)/(dx))`

`(dy)/(dx) = 1/(sec^2y)`

`= 1/(1+tan^2y)`

Since `tany = x`, `tan^2y = x^2`.

`=> d/(dx)[arctanx] = 1/(1 + x^2)`

Therefore, if you recall that `int (dy)/(dx) = y`, then you should find that you have

`color(blue)(int_(0)^(oo) 1/(1+x^2)dx)`

`= |[arctanx]|_(0)^(oo)`

`= lim_(x->oo) arctanx - lim_(x->0) arctanx`

If you think about it, what happens as `x->oo` for `arctanx`? If `arctan(oo) = u`, then `oo = tanu`. `tanu` approaches `oo` at `pm pi/2`. But since we are in `[0,oo)`, we must be talking about `pi/2`.

Therefore, `lim_(x->oo) arctanx = pi/2`.

Further, for `arctan(0) = u`, `0 = tanu = (sinu)/(cosu)`. When `u = 0`, it is sure that `sinu = 0` and `cosu ne 0`. Inverting the graph of `tanu` yields a graph with asymptotes at `pm pi/2` and which intersects `(0,0)`.

Thus, `lim_(x->0) arctanx = 0`.

Now we'd get:

`=> pi/2 - 0`

`= color(blue)(pi/2)`