**15.** Let \( f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x \), \( I_1 = \int_{0}^{\frac{\pi}{4}} f(x) \, dx \) and \( I_2 = \int_{0}^{\frac{\pi}{4}} x f(x) \, dx \). Then \( 7I_1 + 12I_2 \) is equal to: - (1) \( 2\pi \) - (2) \( \pi \) - (3) 1 - (4) 2

1. Understanding the integrand

The given function

contains only even powers of $\tan x$. Setting $t = \tan x$ (so $dt = \sec^2 x\,dx = (1+t^2)\,dx$) is therefore a natural substitution, because it converts powers of $\tan x$ into plain powers of $t$ and replaces the mysterious $dx$ by $\frac{dt}{1+t^2}$.

2. Converting I1

With $t = \tan x$, the limits change from $x=0$ to $t=0$ and from $x=\pi/4$ to $t=1$.

Long-division by $(1+t^2)$ is the crucial trick:

That turns I1 into a plain polynomial integral—quick to evaluate.

3. Converting I2

For

exactly the same substitution gives

The new obstacle is the $\arctan t$. The classic remedy is integration by parts:

• pick $u = \arctan t$ (simple derivative) and $dv = (7t^6-3t^2)\,dt$ (easy antiderivative),
• then $du=\frac{dt}{1+t^2}$ and $v = t^7 - t^3$. Because the $uv$ term drops out neatly at both limits, the remaining integral once again reduces to a polynomial.

4. Final combination

After the two clean evaluations, you form the linear combination $7I_1+12I_2$: I1 turns out to be 0, I2 a tiny fraction, so the mix simplifies to the integer 1.

The whole exercise reinforces three key JEE-level skills:

1. Spotting symmetry/cancellation in rational functions of $\tan x$.
2. Executing a t = tan x substitution confidently.
3. Knowing when and how to invoke integration by parts to tame an extra factor of $x$ (or $\arctan t$).

What’s going on here?

Imagine you have a strange roller-coaster track whose height at every point $x$ is given by a messy formula full of $\tan x$’s.
You first want to know how much total ‘area’ (call it I1) is under that track from the starting gate ($x=0$) up to the point $x=\pi/4$ (that’s 45°).
Then you want a slightly different thing, I2, where you multiply each little strip of area by its x-coordinate before adding it up (so strips farther away count a bit more).
Finally, your teacher mixes those two answers in the recipe 7 I1 + 12 I2 and asks: what number pops out?

How can we attack it?

1. Swap the angle for its slope: The roller-coaster’s steepness is $\tan x$. Replacing $x$ with $t=\tan x$ turns all those powers of $\tan x$ into simple powers of $t$.
2. Notice a magic cancellation: After the swap, the horrible fraction actually collapses to a much simpler polynomial. That makes I1 super easy (it becomes 0!).
3. Use a clever handshake (integration by parts): For I2, the swap gives you an extra $\arctan t$. A standard trick called “integration by parts” cleverly moves that $\arctan$ out of the way.
4. Add, stir, and taste: Crunch the numbers; everything simplifies to the neat little number 1. That is exactly option (3).

So the roller-coaster question secretly hides a very neat cancellation and only wants you to know a couple of standard calculus tools!

Step-by-step calculation

1. Evaluate $I_1$

Substitute $t = \tan x$, so $dx = \frac{dt}{1+t^2}$ and limits become $0 \to 1$.

Long-divide by $1+t^2$:

Hence

2. Evaluate $I_2$

With the same substitution,

Take $u = \arctan t\;\Rightarrow\;du = \dfrac{dt}{1+t^2}$,
$dv = (7t^6 - 3t^2)\,dt\;\Rightarrow\;v = t^7 - t^3$.

Integration by parts:

The boundary term vanishes because $(t^7 - t^3)=0$ at both $t=0$ and $t=1$.

Divide once more:

Therefore

3. Combine as required

[ \boxed{1}\quad\text{(option 3)} ]