You should not drive the LED in the opto-coupler at its absolute max continuous current of 50mA. Then it will also be at its absolute max allowed temperature if its ambient temperature is 25 degrees C.
How did you calculate a current of 43mA with a 82 ohms current-limiting resistor? The darlington driver (which one?) IC has an output voltage loss of about 0.75V and you don't know the exact forward voltage of the IR LED. The datasheet says the forward voltage is a max of 1.5V but it is probably less. What is its typical voltage?
Assuming that the forward voltage is 1.3V at 43mA then the voltage across the current-limiting resistor is 5.0V - 1.3V - 0.75V= 2.95V. Then for 43mA the resistor value is 2.95V/43mA= 68.6 ohms.