Question

Let $f$ be a function defined in $[0,\pi ]$ satisfies the condition
${\int }_0^x\left[f^{\prime }\right(t)-\sin 2t]dt={\int }_x^0\left[f\right(t\left)\tan t\right]dt$
If the curve passes through $(0,2)$
Then the maximum value of $f\left(x\right)$ is

Answer 1

${\int }_0^x\left[f^{\prime }\right(t)-\sin 2t]dt={\int }_x^0\left[f\right(t\left)\tan t\right]dt$
differentiating both side (assuming $y=f\left(x\right)$)
$\therefore f^{\prime }\left(x\right)-\sin 2x=-f\left(x\right)\tan x{f}^{\prime }\left(x\right)+f\left(x\right)\tan x=\sin 2x$
this is a linear differential equation
$\therefore I.F.=e^{\int \tan xdx}=\sec x\Rightarrow y\sec x=\int \sec x\sin 2xdxy=-2{\cos }^{2}x+C\cos x$
$\because$ curve passes through $(0,2)$
$\therefore C=4$
$y=2-2(1+\cos x{)}^2$
$y_{max}=2$