Function Posted February 19, 2014 Share Posted February 19, 2014 (edited) Hello In Math class, we were to sole an integration. While my teacher was explaining and resolving the exercise, I did it myself, finding [math]\int{\cdots}=-\arctan{e^{-x}}=f_1(x)[/math] Later, my teacher found [math]\int{\cdots}=\arctan{e^{x}}=f_2(x)[/math]. As she didn't find any mistake in my work, we agreed that the 'problem' MUST be that those two primitives are different from each other, from a constant. I now find that [math]f_2(x)-f_1(x)=\arctan{e^{x}}+\arctan{e^{-x}}=\frac{\pi}{2}[/math] And that this is not only true for [math]e[/math], but for every real number. Can this be proven? Thanks. Function [EDIT] I found - pure by luck - a plausible proof: [math]\arctan{a^x}+\arctan{a^{-x}}=\frac{\pi}{2}[/math] [math]\Leftrightarrow \arctan{a^{x}}=\frac{\pi}{2}-\arctan{a^{-x}}[/math] [math]\Leftrightarrow a^x= \tan{\left(\frac{\pi}{2}-\arctan{a^{-x}}\right)}[/math] [math]\Leftrightarrow a^x=\frac{\sin{\left(\frac{\pi}{2}-\arctan{a^{-x}}\right)}}{\cos{\left(\frac{\pi}{2}-\arctan{a^{-x}}\right)}}[/math] [math]\Leftrightarrow a^x=\frac{\cos{\arctan{a^{-x}}}}{\sin{\arctan{a^{-x}}}}[/math] [math]\Leftrightarrow a^x=\frac{1}{\tan{\left(\arctan{a^{-x}}\right)}}[/math] [math]\Leftrightarrow \frac{1}{a^x}=\tan{\arctan{\frac{1}{a^x}}}[/math] [math]\Leftrightarrow \frac{1}{a^x}=\frac{1}{a^x}[/math] True [math]\Leftrightarrow \arctan{a^x}+\arctan{a^{-x}}=\frac{\pi}{2}[/math] Q.E.D. Edited February 19, 2014 by Function 2 Link to comment Share on other sites More sharing options...
studiot Posted February 19, 2014 Share Posted February 19, 2014 Well done for doing it yourself. +1 1 Link to comment Share on other sites More sharing options...
mathematic Posted February 20, 2014 Share Posted February 20, 2014 Essentially you have arctan(u) + arctan(1/u) = π/2. If you look at any right triangle, the two acute angles have the relationship. Link to comment Share on other sites More sharing options...
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