Let f(x)=∫0x tsin⁡1tdt. Then the number of points of discontinuity of the  function f(x) in the open interval (0,π) is

f(x)=xsin⁡1x. At all points in (0,π),f′(x) has a definite finite value. ∴ f(x) is differentiable finitely in (0,π). As a finitely differentiable  function is also continuous, hence f(x) is continuous in (0,π).