The value of $\int\limits_0^2 {\frac{{dx}}{{{{(1 - x)}^2}}}} $ is

Let $f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x$ for $x \in R$ be a function which satisfies $f(x)=x+\int \limits_0^{\pi / 2} \sin (x+y) f(y) d y$. Then $(a+b)$ is equal to $............$

$\int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$ is equal to (where $\{.\}$ denotes the fractional part function and $n, r \in N$)

An extremum value of $y = \int_{0}^{x} (t - 1)(t - 2) dt$ is

Let $f(x)$ be a function satisfying $f'(x) = f(x)$ with $f(0) = 1$ and $g(x)$ be the function satisfying $f(x) + g(x) = x^2$. The value of the integral $\int_0^1 f(x)g(x) dx$ is equal to

The values of $\alpha$ which satisfy $\int_{\pi /2}^{\alpha} \sin x \, dx = \sin 2\alpha$,where $\alpha \in [0, 2\pi]$,are equal to: