\(\displaystyle{\int_{{x}}^{{1}}}{\frac{{{\left.{d}{t}\right.}}}{{{1}+{t}^{{2}}}}}={\int_{{1}}^{{\frac{{{1}}}{{{x}}}}}}{\frac{{{\left.{d}{t}\right.}}}{{{1}+{t}^{{2}}}}}\) I showed that \(\displaystyle{\int_{{x}}^{{1}}}{\frac{{{\left.{d}{t}\right.}}}{{{1}+{t}^{{2}}}}}-{\int_{{1}}^{{\frac{{{1}}}{{{x}}}}}}{\frac{{{\left.{d}{t}\right.}}}{{{1}+{t}^{{2}}}}}={\frac{{\pi}}{{{4}}}}-{\arctan{{\left({x}\right)}}}-{\left[{\arctan{{\left({\frac{{{1}}}{{{x}}}}\right)}}}-{\frac{{\pi}}{{{4}}}}\right]}\) Since \(\displaystyle{\arctan{{\left({x}\right)}}}+{\arctan{{\left({\frac{{{1}}}{{{x}}}}\right)}}}={\frac{{\pi}}{{{2}}}}\), then

Question

Polchinio3es

Answered question

2022-03-31

\int_{x}^{1} \frac{d t}{1 + t^{2}} = \int_{1}^{\frac{1}{x}} \frac{d t}{1 + t^{2}}

I showed that

\int_{x}^{1} \frac{d t}{1 + t^{2}} - \int_{1}^{\frac{1}{x}} \frac{d t}{1 + t^{2}} = \frac{π}{4} - \arctan (x) - [\arctan (\frac{1}{x}) - \frac{π}{4}]

Since

\arctan (x) + \arctan (\frac{1}{x}) = \frac{π}{2}

, then it follows that

\int_{x}^{1} \frac{d t}{1 + t^{2}} - \int_{1}^{\frac{1}{x}} \frac{d t}{1 + t^{2}} = 0

. However, I feel this is kind of proof only "shows" but doesn't really "explains".
Is there another way to prove this? The simpler the better!