If $I_1 = \int_0^{\pi / 2} \frac{x}{\sin x} dx$ and $I_2 = \int_0^1 \frac{\tan^{-1} x}{x} dx$,then $I_1 : I_2$ is

$A$ continuous and differentiable function $f$ satisfies the condition $\int_{0}^{x} f(t) dt = f^2(x) - 1$ for all real $x$. Then:

Value of the $\int_0^9 \sqrt{x} \,dx + \int_0^{\pi/2} (\cos x + \sin x) \,dx$ is:

Let $u = \int_{0}^{\infty} \frac{dx}{x^4 + 7x^2 + 1}$ and $v = \int_{0}^{\infty} \frac{x^2 dx}{x^4 + 7x^2 + 1}$. Then:

Let $f: R \rightarrow R$ be a differentiable function such that its derivative $f^{\prime}$ is continuous and $f(\pi)=-6$. If $F:[0, \pi] \rightarrow R$ is defined by $F(x)=\int_0^{ x } f( t ) dt$,and if $\int_0^\pi\left(f^{\prime}( x )+ F ( x )\right) \cos x dx =2$,then the value of $f(0)$ is.

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: