Animated Solution for Mathematics - Definite Integration: If $(a,b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $I_1={\int }_a^{b}x\sin (4x-x^2)dx,I_2={\int }_a^b\sin (4x-x^2)dx$, then $36\frac{I_1}{I_2}$ is equal to :

Welcome, student. Today, we are not just solving a problem; we are embarking on a journey that bridges the gap between the rigid lines of coordinate geometry and the fluid, beautiful world of calculus. This problem is a classic JEE Advanced masterpiece.

It tests your ability to maintain composure when the geometry seems tedious and your insight when the calculus seems impossible. Let us begin.

We start with a triangle defined by vertices $A(1,2)$, $B(2,3)$, and $C(3,1)$. Our first objective is to find the orthocenter $H(a,b)$. Remember, the orthocenter is the point where the altitudes meet.

First, consider the altitude from $A$ to $BC$. The slope of $BC$ is calculated as:

Since the altitude is perpendicular to $BC$, its slope must be the negative reciprocal, which is $\frac{1}{2}$. Using the point-slope form at $A(1,2)$, the equation becomes $(y-2)=\frac{1}{2}(x-1)$, which simplifies to:

Next, we find the altitude from $B$ to $AC$. The slope of $AC$ is $m_{AC}=\frac{1-2}{3-1}=-\frac{1}{2}$. The altitude is perpendicular, so its slope is $2$. Using vertex $B(2,3)$, we get $(y-3)=2(x-2)$, which simplifies to:

Now, we solve these two linear equations simultaneously. Substituting $y=2x-1$ into the first equation, we find:

Consequently, $y=2(\frac{5}{3})-1=\frac{7}{3}$. So, our orthocenter is $H(\frac{5}{3},\frac{7}{3})$.

The problem asks us to integrate from $a$ to $b$. Let us calculate their sum:

This is the 'Aha!' moment. The sum of the limits is $4$, which is the key that unlocks the entire integral.

We are faced with $I_1={\int }_a^{b}x\sin (4x-x^2)dx$ and $I_2={\int }_a^b\sin (4x-x^2)dx$. When you see limits $a$ and $b$ and a function that looks difficult to integrate directly, you must summon the King's Rule:

Since $a+b=4$, we replace $x$ with $(4-x)$ in $I_1$. Watch what happens to the argument of the sine function:

The argument remains unchanged! This is the elegance of the problem.

Now, apply this to $I_1$:

We can split this into two integrals:

Notice that the first part is exactly $4I_2$, and the second part is our original $I_1$. Thus, we have the relation $I_1=4I_2-I_1$. Rearranging this gives $2I_1=4I_2$, or:

Finally, the problem asks for $36\frac{I_1}{I_2}$. Substituting our result, we get $36\times 2=72$.

The final answer is 72.