Where's the mistake?

Question

∫1/(cos² x + sin² x) dx from ∏ to 0 =
∫sec² x/(1+tan² x) dx = arctan (tan x) |∏ to 0 = 0

i know the mistake is with arctan (tan x) |∏ to 0 , but i cannot show why, as
∫f'(x)/(a²+(f(x))² dx = 1/a arctan[(f(x)/a]....can someone enlighten me? thanks

Answer 1

I guess i recognized your mistake actually you forget the principle that you are allowed to put the integral bounds in the expresion if just the expresion is continues in that range here tan(x) is not continues in range 0 to pi (at the point pi/2)
thus you can solve the problem in tow ways first you can divide the interval into two parts in which tan(x) is continues in both and then sum the results :
arctan(tan(x))|0 to pi=arctan(tan(x))|0 to pi/2- + arctan(tan(x))|pi/2+ to pi
then as you know tan(pi/2-)=+infinite and arctan(+infinite)=pi/2 tan(pi/2-)=-infinite and arctan(-infinite)=-pi/2 as you you see the sum of results equals pi

the second way is more simpler you know arctan(tan(x))=x :) then you just need calculate x|0 to pi=pi :)

hope you understand well

Answer 2

like Helmut said, your mistake is not recognizing that (sin^2 x + cos^2 x)=1.


At any rate, in a different but related question, if you want to know why ∫f'(x)/(a²+(f(x))² dx = 1/a arctan[(f(x)/a] the answer is below.

You go about finding what the derivative of tan ‾ ¹(x) is then you know what integrates to give you tan ‾ ¹(x).


tanï´¾tan ‾ ¹(x)ï´¿ = x for all x

let y = tan ‾ ¹(x)

so we have:

tan(y) = x

now differentiate both sides (you need to use implicit differentiation)

sec²(y)dy/dx = 1

dy/dx = 1/ sec²(y)

by trig. Identity:

dy/dx = 1/ ﴾ 1 + tan²(y) ﴿

But look above, y = tan ‾ ¹(x), so tan²(y) = tan²( tan ‾ ¹(x)) = x²

Thus

dy/dx = 1/ ﴾ 1 + x²﴿


the constant ‘a’ and the chain rule are trivially added to the proof.

Answer 3

To go from...∫1/(cos² x + sin² x) dx....to ∫sec² x/(1+tan² x) dx, you are multiplying all quantities by 1/cos² x, which means that your equations are not differentiable when cos x = 0, notably at pi/2.

You are also neglecting the "practical periodicity" of the arctan function. When you are talking about the actual graph, arctan(tan(0)) and arctan(tan(pi)) should be equal to different things.

This is analogous (at least to me, in the quirky nature of my mind) like losing a root when taking the square root of both sides of an equation.

Answer 4

I think your mistake is in not recognizing that (sin^2 x + cos^2 x)=1 , and so the integral becomes trivial, having a value of -PI.