Could someone explain me this "ownership" of the arctangent someone could explain to me this: 
$$\int { \arctan { \left( \frac { 1 }{ { u }^{ 2 } }  \right)  }  }  \,du=\int { \frac { \pi  }{ 2 }  } -\arctan { \left( { u }^{ 2 } \right)  }  \, du$$

$$\boxed{\tan^{-1} \dfrac 1 x = \cot^{-1}x = \dfrac {\pi} 2 - \tan^{-1} x}$$
Proof of first equality: $y = \tan^{-1} \dfrac 1 x \Rightarrow \tan y = \dfrac 1 x \Rightarrow \cot y = x \Rightarrow y = \cot^{-1} x$.
Proof of second equality: $\cot \left(\dfrac {\pi} 2 - \theta\right) = \tan \theta \Rightarrow \cot \left(\dfrac {\pi} 2 - \tan^{-1} x \right) = \tan \tan^{-1}x = x \Rightarrow \dfrac {\pi} 2 - \tan^{-1} x = \cot^{-1} x$.
