Equations on Integration!

$\large{\int_0^x t \cdot f(x-t)\ \mathrm{d}t = \int_0^x f(t)\ \mathrm{d}t + \sin(x) + \cos(x) -x-1}$ $\color{#3D99F6}{\text{As} \int_a^b f(x) \mathrm{d}t= \int_a^b f((a+b)-x)\mathrm{d}t}$

$\large{\int_0^x (x-t) \cdot f(t)\ \mathrm{d}t = \int_0^x f(t)\ \mathrm{d}t + \sin(x) + \cos(x) -x-1}$

$(x-1)\int_0^x f(t)\ \mathrm{d}t - \large{\int_0^x t \cdot f(t)\ \mathrm{d}t = \sin(x) + \cos(x) -x-1}$

Differentiate above equation with respect to $x$ .

$(x-1) \cdot f(x) + \int_0^x f(t) \mathrm{d}t - x \cdot f(x) = \cos(x)-\sin(x) -1$

Subtacting $x.f(x)$ on the LHS. And rearranging the equation.

$\int_0^x f(t) \mathrm{d}t =f(x)+\cos(x)-\sin(x) -1$

Again differentiate the above equation.

$f(x) =f'(x)-\cos(x)-\sin(x)$

Above equation is a First Order Linear Differential Equation which can be easily solved. And finally we get:

$f(x) = \color{#20A900}{c} \cdot e^x - \cos(x)$

Here $\color{#20A900}{c}$ is an arbitrary constant. Value of the constant can be found out by using above equations which will turn out to be $1$ .

$\boxed{ f(x) = e^x - \cos(x)}$

Putting $x = \frac{π}{6}$ we get: $\boxed{f(\frac{\pi}{6})=e^{\frac{\pi}{6}}-\frac{√3}{2}=0.822}$