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, 201411 yr by Function

Well done for doing it yourself. +1

Essentially you have arctan(u) + arctan(1/u) = π/2. If you look at any right triangle, the two acute angles have the relationship.