Animated Solution for Mathematics - Definite Integration: Let $f(x)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x$ for all $x\in (-\frac{\pi }{2},\frac{\pi }{2})$. Then the correct expression(s) is(are)

Imagine you are standing before a massive, complex-looking function:

It looks like a monster, doesn't it? A chaotic collection of high-power trigonometric terms that seem designed to make your head spin.

But here is the secret of JEE Advanced physics and mathematics: the most intimidating problems are often the ones that hide the most elegant solutions. Today, we are going to dismantle this monster, piece by piece, and find the beauty hidden within.

When you first see an expression like this, your instinct might be to panic or to try some brute-force integration. Resist that urge! Instead, take a deep breath and look for patterns.

Mathematics is the study of patterns, and this function is practically screaming for us to group its terms. Look at the first two terms: $7{\tan }^{8}x+7{\tan }^{6}x$. We can factor out $7{\tan }^{6}x$ to get $7{\tan }^{6}x({\tan }^{2}x+1)$.

Now look at the last two terms: $-3{\tan }^{4}x-3{\tan }^{2}x$. We can factor out $-3{\tan }^{2}x$ to get $-3{\tan }^{2}x({\tan }^{2}x+1)$.

Do you see it now? The common factor $({\tan }^{2}x+1)$ has emerged like a lighthouse in the fog! Our function is simply:

Now, we invoke one of the most powerful tools in our trigonometric arsenal: the identity $1+{\tan }^{2}x={\sec }^{2}x$. By substituting this into our factored expression, the function transforms into:

Suddenly, the integration becomes a walk in the park. We want to evaluate $\int f(x)dx$. With the ${\sec }^{2}x$ term sitting there, we know that if we let $u=\tan x$, then $du={\sec }^{2}xdx$.

The integral becomes $\int (7u^6-3u^2)du$. This is just a simple polynomial! Integrating this gives us $u^7-u^3$.

Substituting back, we get ${\tan }^{7}x-{\tan }^{3}x$. When we evaluate this from $0$ to $\frac{\pi }{4}$, we get $(1-1)-(0-0)=0$. The net area is zero!

But we are not done yet. The problem also asks us to evaluate ${\int }_0^{\frac{\pi }{4}}xf(x)dx$. Now we have a product of two functions: $x$ and $f(x)$.

This is the classic setup for Integration by Parts. We choose $u=x$ and $dv=f(x)dx$. The formula $\int udv=uv-\int vdu$ becomes our roadmap.

We already know $v=\int f(x)dx={\tan }^{7}x-{\tan }^{3}x$. So, our integral is:

As we discussed, the boundary term vanishes completely because $\tan (\frac{\pi }{4})=1$. We are left with $-{\int }_0^{\frac{\pi }{4}}({\tan }^{7}x-{\tan }^{3}x)dx$, which is ${\int }_0^{\frac{\pi }{4}}({\tan }^{3}x-{\tan }^{7}x)dx$.

We are in the home stretch now. We can factor this as:

Using the difference of squares, this becomes:

Again, we see that $1+{\tan }^{2}x$ term, which is ${\sec }^{2}x$. With another substitution $t=\tan x$, the integral becomes:

Evaluating this, we get:

We have tamed the monster! We have navigated the complexity, used our tools wisely, and arrived at the elegant answer of $\frac{1}{12}$. Remember, in JEE Advanced, it is not about memorizing formulas; it is about seeing the structure and trusting the process.