If x≥1, then 2tan−1x+sin−12x1+x2 is equal to
4tan−1x
0
π/2
π
We have,
sin−12x1+x2=π−2tan−1x for x≥1
∴ 2tan−1x+sin−12x1+x2=2lan−1x+π−2tan−1x=π