Op Amp integrator (square to triangle) not always start working correctly
The condition for integration using differential equations are;

Ic =C dV/dt  where dV = output Vpp in dt ramp time for 1/2 triangle time    


Thus if you wanted to integrate a 1Vp 1kHz square wave into a 5Vpp Triangle ( using a RRIO OA ) 
dV/dt = 10V/ms from  5V for each half cycle of 0.5ms  
Ic= 100 uA = Vin/R = 1Vp /10k  (Pot max) 
thus C = Ic* dt  /dV   = 10 nF  [uA*ms/V] 


Whereas you show 10uF which is far too big as pots only have a 300:1 range and 10 Ohms is too high a current to drive with 10uF @ 1kHz.   

Always use specs and 1st principles when in doubt.  Ic=CdV/dt

For a better design with included sig Gen.

Use a Schmitt Trigger Astable that buffers the Triangle wave to constant amplitude.