If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $I_1={\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{xsin}\left(4\mathrm{x}-{\mathrm{x}}^2\right)\mathrm{dx},{\mathrm{I}}_2={\int }_{\mathrm{a}}^{\mathrm{b}}\sin \left(4\mathrm{x}-{\mathrm{x}}^2\right)\mathrm{dx}$ , then $36\frac{I_1}{I_2}$ is equal to :

Equation of $\mathrm{CE}$ is given by,

$\Rightarrow \left(y-1\right)=\frac{-1}{{\displaystyle \frac{3-2}{2-1}}}\left(x-3\right)$

$\Rightarrow \left(y-1\right)=-(x-3)$

$\Rightarrow x+y=4 ...\left(i\right)$

Similarly, equation of $AD$ is given by,

$\Rightarrow \left(y-2\right)=\frac{-1}{{\displaystyle \frac{3-1}{2-3}}}\left(x-1\right)$

$\Rightarrow \left(y-2\right)=\frac{1}{{\displaystyle 2}}\left(x-1\right)$

$\Rightarrow 2y-4=x-1$

$\Rightarrow x-2y+3=0$

Using $\left(i\right)$

$\Rightarrow 4-y-2y+3=0$

$\Rightarrow y=\frac{7}{3}$

$\Rightarrow x=4-\frac{7}{3}$

$\Rightarrow x=\frac{5}{3}$

So, Orthocentre is $H\equiv \left(a,b\right)=\left(\frac{5}{3},\frac{7}{3}\right)$.

Orthocentre lies on the line $x+y=4$

So, $a+b=4$

Now, $I_1={\int }_a^{b}x\sin \left(x\right(4-x\left)\right)dx ...\left(1\right)$

$\Rightarrow I_1={\int }_a^b(a+b-x)\sin \left(\left(a+b-x\right)\right(4-\left(a+b-x\right)\left)\right)dx$

$\Rightarrow I_1={\int }_a^b(4-x)\sin \left(x\right(4-x\left)\right)dx ...\left(2\right)$

Using, $\left(1\right) + \left(2\right)$

$\Rightarrow 2I_1={\int }_a^{b}4\sin \left(x\right(4-x\left)\right)dx$

$\Rightarrow 2{\mathrm{I}}_1=4{\mathrm{I}}_2$

$\Rightarrow {\mathrm{I}}_1=2{\mathrm{I}}_2$

$\Rightarrow \frac{I_1}{I_2}=2$

$\Rightarrow \frac{36I_1}{I_2}=72$