If $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a continuous function satisfying $$\int_\limits{0}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x+\alpha \int_\limits{0}^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x=0$$, then the value of $$\alpha$$ is :

Let the function $$f:[0,2] \rightarrow \mathbb{R}$$ be defined as

$$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$$

where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_\limits{0}^{2} x f(x) d x$$ is :

The value of the integral $$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$ is equal to :

Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then $$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to :