1
$\begingroup$

someone could explain to me this:

$$\int { \arctan { \left( \frac { 1 }{ { u }^{ 2 } } \right) } } \,du=\int { \frac { \pi }{ 2 } } -\arctan { \left( { u }^{ 2 } \right) } \, du$$

$\endgroup$
  • $\begingroup$ See this or this. $\endgroup$ – David Mitra Jun 26 '14 at 17:15
  • $\begingroup$ What is an "ownership"? $\endgroup$ – David K Jun 26 '14 at 19:30
  • $\begingroup$ I voted to put this on hold because it is not clear to me what is being asked. $\endgroup$ – Carl Mummert Jun 26 '14 at 20:36
3
$\begingroup$

If you draw a right triangle with sides $a,b,c$ angle $A$ opposite leg $a$, you have $\arctan \frac ab=A, \arctan \frac ba=\frac \pi 2-A$. The general relation then is $\arctan \frac 1x=\frac \pi 2 -\arctan x$. The integral signs don't matter.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

$$\boxed{\tan^{-1} \dfrac 1 x = \cot^{-1}x = \dfrac {\pi} 2 - \tan^{-1} x}$$

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: $\cot \left(\dfrac {\pi} 2 - \theta\right) = \tan \theta \Rightarrow \cot \left(\dfrac {\pi} 2 - \tan^{-1} x \right) = \tan \tan^{-1}x = x \Rightarrow \dfrac {\pi} 2 - \tan^{-1} x = \cot^{-1} x$.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.