$$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{\cos x} \over {1 + {e^x}}}dx = } $$

$$\int_\limits0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1-\sin x \cos x} d x=$$

If $$f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$$, then $$\int_\limits1^4 f(x) d x=$$

If $$2 f(x)-3 f\left(\frac{1}{x}\right)=x$$, then $$\int_\limits1^e f(x) d x=$$