If $$b = \int\limits_0^1 {{{{e^t}} \over {t + 1}}dt} $$, then $$\int\limits_{a - 1}^a {{{{e^{ - t}}} \over {t - a - 1}}} $$ is

Let $$I = \int_{\pi /4}^{\pi /3} {{{\sin x} \over x}dx} $$. Then

The value of

$$\sum\limits_{n = 1}^{10} {} \int\limits_{ - 2n - 1}^{ - 2n} {{{\sin }^{27}}} x\,dx + \sum\limits_{n = 1}^{10} {} \int\limits_{2n}^{2n + 1} {{{\sin }^{27}}} x\,dx$$ is equal to

$$\int\limits_0^2 {[{x^2}]} \,dx$$ is equal to