The integral $$\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$$ equals :

$$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$

If I₁ = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$$ and
I₂ = $$\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$$ such
that I₂ = $$\alpha $$I₁ then $$\alpha $$ equals to :

The value of $$\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $$ is: