<math>\sqrt{\frac{1+2u^2+u^4-(1-2u^2+u^4)}{(1+u^2)^2}}=\sqrt{\frac{4u^2}{(1+u^2)^2}}=\sqrt{\frac{(2u)^2}{(1+u^2)^2}}=\frac{2u}{1+u^2}</math>
ובדרך אחרת:
<math>\tan(\frac{x}{2})=\frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})}=\frac{2 \cdot \sin(\frac{x}{2}) \cdot \cos(\frac{x}{2})}{2 \cos^2(\frac{x}{2})}=\frac{\sin(x)}{2 \cos^2(\frac{x}{2})}</math>
ולכן מתקיים
<math>\sin(x)=\tan(\frac{x}{2})\cdot 2 \cos^2(\frac{x}{2})=\frac{2u}{1+u^2}</math>
כמו כן, <math>x=2\arctan(u)\ \Rightarrow\ dx=\frac{2}{1+u^2}du</math> .