One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in x=0 x = 0 .
Here's another way. Look at π2−2arctanx−−√=2(π4−arctanx−−√)=2(arctan1−arctanx−−√). π 2 − 2 arctan x = 2 ( π 4 − arctan x ) = 2 ( arctan 1 − arctan x ) . Now remember the identity for the difference of two arctangents: arctanu−arctanv=arctanu−v1+uv. arctan u − arctan v = arctan u − v 1 + u v . (This follows from the usual identity for the tangent of a sum.) The left side above becomes 2arctan1−x−−√1+x−−√. 2 arctan 1 − x 1 + x . The double-angle formula for the sine says sin(2u)=2sinucosu sin ( 2 u ) = 2 sin u cos u . Apply that: sin(2arctan1−x−−√1+x−−√)=2sin(arctan1−x−−√1+x−−√)cos(arctan1−x−−√1+x−−√) sin ( 2 arctan 1 − x 1 + x ) = 2 sin ( arctan 1 − x 1 + x ) cos ( arctan 1 − x 1 + x ) Now remember that sin(arctanu)=u1+u2−−−−−√ sin ( arctan u ) = u 1 + u 2 and cos(arctanu)=11+u2−−−−−√ cos ( arctan u ) = 1 1 + u 2 Then use algebra: 2⋅(1−x√1+x√)1+(1−x√1+x√)2−−−−−−−−−−√⋅11+(1−x√1+x√)2−−−−−−−−−−√=1−x1+x.
