<s>[INST] Proving that $F(x)$ is a constant This was on a test and i know i was supposed to use 2nd ftoc to prove that $F(x)$ was a constant when $x>0$
$$
F(x) = \int_{0}^{x} \frac{1}{t^2 +1} dt + \int_{0}^{\frac{1}{x}} \frac{1}{t^2 +1} \therefore
$$
knowing the following:
$$
F(x) = \int_{a}^{x} f(t) dt
$$
then
$$
F'(x) = f(x)
$$
I would simply do
$$
F'(x) = \frac{1}{x^2+1} + \frac{1}{\frac{1}{x^2}+1} = \frac{1}{x^2+1}+\frac{x^2}{x^2+1} = 1
$$
But couldn't i do this without knowing the 2nd ftoc?
Example:
Set t = tan(x), which then makes the bounds change
$$
\int_{0}^{\arctan(x)}\cos^2(t)dt + \int_{0}^{\arctan(\frac{1}{x})}\cos^2(t)dt
$$
$$
= \frac{1}{2}(\arctan(x)+\sin(\arctan(x))\cos(\arctan(x)) + \arctan(\frac{1}{x})+\sin(\arctan(\frac{1}{x}))\cos(\arctan(\frac{1}{x}))
$$
then use the arctan identities and knowing $\arctan(x)+\arctan(\frac{1}{x}) = \frac{\pi}{2}$ when $x>0$ and do:
$$
\frac{1}{2}[\frac{\pi}{2} + \frac{1}{\sqrt{x^2+1}} \times\frac{x}{\sqrt{x^2+1}} + \frac{1}{x\sqrt{x^{-2}+1}} \times \frac{1}{\sqrt{x^{-2}+1}}]
$$
which equals:
$$
\frac{1}{2}[\frac{\pi}{2} + \frac{x}{x^2+1} + \frac{1}{\frac{1}{x}+x}]
$$
which is not a constant, Where did i go wrong?
 [/INST] You forgot to replace the $dt$ with the proper expression in the substitution. With $t = \tan x$, we have $dt = \tan' x\,dx = (1+\tan^2 x)\,dx$, so the integral becomes
$$\int_0^{\arctan x} \frac{1+\tan^2 x}{1+\tan^2 x}\,dx = \int_0^{\arctan x} 1\,dx = \arctan x$$
and analogously for the other integral.
</s>