Evaluate ∫010π tan−1xdx, where [⋅] represents greatest integer function.
10π
10π-tan1
10π+tan1
10 π + tan 2
Here, y=tan−1x is a monotonic function.
We have 0<tan−1x<1; when 0<x<tan11<tan−1x<π2; when tan1<x<10π
∴ I=∫010π tan−1xdx=∫0tan1 0dx+∫tan110π 1dx
=10π−tan1