Animated Solution for Mathematics - Definite Integration: Let $f(x)=\left|\begin{array}{ccc}\sec x& \cos x& {\sec }^{2}x+\cot x\csc x\\ {\cos }^{2}x& {\cos }^{2}x& {\csc }^{2}x\\ 1& {\cos }^{2}x& {\cos }^{2}x\end{array}\right|$. Then ${\int }_0^{\pi /2}f(x)dx=\dots \dots \dots$

Welcome, future IITians! Today, we are not just solving a problem; we are performing surgery on a mathematical expression.

When you first glance at the function

your instinct might be to panic. It looks like a chaotic mess of trigonometric identities, but remember, in the JEE Advanced arena, complexity is often a mask for simplicity. Our goal is to peel back that mask.

The Surgical Strike

If you try to expand this determinant directly, you will find yourself drowning in a sea of trigonometric terms. That is the trap. The examiner wants to see if you can spot the structure.

Look at the first column: $\sec x$ in the first row and $1$ in the third row. This is a golden opportunity. By applying the row operation $R_1\to R_1-\sec x\cdot R_3$, we can create a zero in the first position.

But wait, look at the second column! $\cos x$ and ${\cos }^{2}x$. When we perform the same operation, $\cos x-\sec x\cdot {\cos }^{2}x$ becomes $\cos x-(\frac{1}{\cos x})\cdot {\cos }^{2}x=\cos x-\cos x=0$. Just like that, we have two zeros in the first row!

The Algebraic Dance

With two zeros in the first row, the determinant collapses. We are left with the third element of the first row multiplied by the minor of the remaining $2\times 2$ matrix.

The minor is

which simplifies to ${\cos }^{4}x-{\cos }^{2}x=-{\cos }^{2}x{\sin }^{2}x$.

When you multiply this by the third element, the expression simplifies beautifully to $f(x)=-{\sin }^{2}x-{\cos }^{5}x$. It is a moment of pure mathematical elegance—the chaos has vanished, leaving behind a clean, integrable function.

The Integration Finale

Now, we face the integral

We use the linearity of the integral to split this into two parts. And here, we bring in our heavy artillery: Wallis' Formula.

For ${\int }_0^{\pi /2}{\sin }^{2}xdx$, the power is even, so we get:

For ${\int }_0^{\pi /2}{\cos }^{5}xdx$, the power is odd, so we get:

Combining these with the negative sign, we arrive at our final answer:

You have conquered the beast! Keep this mindset—look for the structure, simplify, and then execute.