Animated Solution for Mathematics - Definite Integration: Let $f_n={\int }_0^{\pi /2}\left({\sum }_{k=1}^n{\sin }^{k-1}x\right)\left({\sum }_{k=1}^n(2k-1){\sin }^{k-1}x\right)\cos xdx,n\in \mathbb{N}$. Then $f_{21}-f_{20}$ is equal to

Imagine you are sitting in the examination hall. You turn the page, and there it is. A monstrous integral:

Your heart rate spikes. It looks like a chaotic mess of summations, trigonometric powers, and a lingering cosine term. But take a deep breath. In the world of JEE Advanced, intimidation is often the first layer of the problem. The complexity is a mask. Our job is to peel it away.

The First Step of Clarity

Whenever you see a function and its derivative sitting side-by-side, your instincts should scream 'Substitution!' We have $\sin x$ raised to various powers, and we have $\cos xdx$ sitting right there at the end. This is a gift.

Let us define $t=\sin x$. Immediately, the differential becomes $dt=\cos xdx$.

Now, look at the limits. When $x=0$, $t=0$. When $x=\pi /2$, $t=1$. The integral transforms into something much cleaner:

We have successfully moved from the world of trigonometry to the world of polynomials. This is the first victory.

The Hidden Relationship

Let us label our two series. Let $S_1={\sum }_{k=1}^n{t}^{k-1}$ and $S_2={\sum }_{k=1}^n(2k-1)t^{k-1}$. If you try to sum these using standard geometric progression formulas, you will find yourself drowning in algebraic fractions. There must be a deeper connection.

Look at $S_1=1+t+t^2+\cdots +t^{n-1}$. Now look at $S_2=1+3t+5t^2+\cdots +(2n-1)t^{n-1}$.

Do you see it? The coefficients in $S_2$ are odd numbers, and the powers in $S_1$ are integers. If we multiply $S_1$ by $\sqrt{t}$, we get:

The Masterstroke

Now, differentiate this new variable $z$ with respect to $t$. The derivative of $t^{k-1/2}$ is $(k-1/2)t^{k-3/2}$.

This looks slightly different, but watch what happens when we factor out $\frac{1}{2\sqrt{t}}$. We get:

Suddenly, the entire second series $S_2$ appears inside the parentheses! We have discovered that $dz=\frac{1}{2\sqrt{t}}{S}_{2}dt$. This means $S_{2}dt=2\sqrt{t}dz$. This is the 'Aha!' moment that separates the top rankers from the rest.

The Collapse

Let us substitute this back into our integral. We know $S_1=\frac{z}{\sqrt{t}}$. We know $S_{2}dt=2\sqrt{t}dz$.

When we multiply them, the $\sqrt{t}$ terms cancel out perfectly:

The entire complex expression has collapsed into the integral of $2zdz$. We must update our limits one last time. When $t=0$, $z=0$. When $t=1$, $z=1+1+\cdots +1$ ($n$ times), so $z=n$.

The integral becomes:

This is trivial! The result is $f_n=[z^2{]}_0^n=n^2$.

We started with a terrifying integral, and through the power of substitution and pattern recognition, we reduced it to $f_n=n^2$. The question asks for $f_{21}-f_{20}$.

That is simply $21^2-20^2$. Using the identity $a^2-b^2=(a-b)(a+b)$, we get:

The monster is defeated. You didn't just solve a problem; you navigated a labyrinth of logic. Keep this elegance in your toolkit—it is the hallmark of a true mathematician. The final answer is 41.