The option(s) with the values of a and L that satisfy the following equation is(are) (∫₀⁴ᵖⁱ eᵗ (sin⁶

Q: Wait, why does $g(t)$ have a period of $\pi$ when $a$ is even? Shouldn't it be $2\pi$?

That is a fantastic question! Remember that $\sin(at)$ and $\cos(at)$ have a period of $\frac{2\pi}{a}$. If $a$ is an even integer (like $2$ or $4$), then $\frac{2\pi}{a}$ is either $\pi$ or $\frac{\pi}{2}$. Since $\pi$ is a multiple of both, the function $g(t)$ will definitely repeat its values every $\pi$. It is like a wave that completes multiple full cycles within that $\pi$ interval, so the pattern is guaranteed to reset!

Q: Can I just integrate the function directly? It looks like a standard trigonometric integral.

I admire your ambition, but please don't! Integrating $\sin^6(at)$ and $\cos^4(at)$ multiplied by $e^t$ directly would lead to a nightmare of integration by parts or complex reduction formulas. In JEE Advanced, the goal is to spot the symmetry or the periodic structure. The problem is designed to reward the 'smart' path—using the periodicity to cancel out the complex integral entirely.

Q: Why did the integral $\int_0^{\pi} e^t g(t) dt$ cancel out so perfectly?

It feels like magic, doesn't it? Because we showed that every chunk of the integral is just a scaled version of the first chunk (the one from $0$ to $\pi$), we could factor that base integral out of the entire numerator. When you divide the numerator by that same base integral to find $L$, they simply vanish. It is the beauty of mathematical structure—the 'messy' part of the problem was never meant to be solved; it was meant to be eliminated!

Question

The option(s) with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int_0^{4\pi} e^t (\sin^6 at + \cos^4 at) dt}{\int_0^\pi e^t (\sin^6 at + \cos^4 at) dt} = L?$

Options:

(A) $a = 2, L = \frac{e^{4\pi} - 1}{e^\pi - 1}$

(B) $a = 2, L = \frac{e^{4\pi} + 1}{e^\pi + 1}$

(C) $a = 4, L = \frac{e^{4\pi} - 1}{e^\pi - 1}$

(D) $a = 4, L = \frac{e^{4\pi} + 1}{e^\pi + 1}$

Sigma Percentile · Accepted Answer

### The Mystery of the Periodic Integral

Imagine you are standing before a massive, intimidating wall of calculus. You see an integral from $0$ to $4\pi$ involving $e^t$ and a complex trigonometric expression: $g(t) = \sin^6(at) + \cos^4(at)$.

Your first instinct might be to reach for your pen and start grinding through integration by parts, but stop! Take a deep breath. In the world of JEE Advanced, the most complex-looking problems often hide the most elegant, simple solutions.

Today, we are going to peel back the layers of this 'monster' integral and find the beauty hidden underneath.

### Phase 1: The Periodic Insight

First, let's look at our function $g(t) = \sin^6(at) + \cos^4(at)$. The problem gives us a hint by asking about the values of $a$.

What happens if $a$ is an even integer? If $a=2$, the function becomes $g(t) = \sin^6(2t) + \cos^4(2t)$.

The period of $\sin(2t)$ is $\pi$, and the period of $\cos(2t)$ is $\pi$. Therefore, the entire function $g(t)$ repeats itself every $\pi$.

This is our golden key. If $g(t + \pi) = g(t)$, we have unlocked the ability to break this massive integral into manageable pieces.

### Phase 2: The Strategy of Decomposition

We are integrating from $0$ to $4\pi$. Since our function $g(t)$ resets every $\pi$, it makes perfect sense to slice this integral into four equal, bite-sized chunks:

$$I_{num} = \int_0^{\pi} e^t g(t) dt + \int_{\pi}^{2\pi} e^t g(t) dt + \int_{2\pi}^{3\pi} e^t g(t) dt + \int_{3\pi}^{4\pi} e^t g(t) dt$$

Instead of staring at the whole thing, let's focus on a general interval, the $k$-th chunk, from $k\pi$ to $(k+1)\pi$, where $k$ is $0, 1, 2,$ or $3$. This is where the magic happens.

### Phase 3: The Substitution

Let's perform a substitution to shift our limits back to the familiar territory of $0$ to $\pi$. Let $t = u + k\pi$, which implies $dt = du$.

When $t = k\pi$, $u = 0$. When $t = (k+1)\pi$, $u = \pi$. Our integral for the $k$-th chunk becomes:

$$I_k = \int_0^{\pi} e^{u+k\pi} g(u+k\pi) du$$

Using the laws of exponents, $e^{u+k\pi} = e^{k\pi} \cdot e^u$. And because of our periodic insight, $g(u+k\pi) = g(u)$.

So, the integral simplifies to:

$$I_k = e^{k\pi} \int_0^{\pi} e^u g(u) du$$

Do you see it? Every single chunk is just the base integral $\int_0^{\pi} e^u g(u) du$ multiplied by a scaling factor $e^{k\pi}$.

### Phase 4: The Geometric Progression

Now, we sum these chunks up. We factor out the base integral, and we are left with a beautiful geometric series:

$$I_{num} = (e^0 + e^{\pi} + e^{2\pi} + e^{3\pi}) \int_0^{\pi} e^t g(t) dt$$

The sum $1 + e^{\pi} + e^{2\pi} + e^{3\pi}$ is a classic geometric progression with first term $A=1$, common ratio $R=e^{\pi}$, and $n=4$ terms.

Using the sum formula $S_n = \frac{A(R^n - 1)}{R - 1}$, we get:

$$S = \frac{(e^{\pi})^4 - 1}{e^{\pi} - 1} = \frac{e^{4\pi} - 1}{e^{\pi} - 1}$$

### The Grand Finale

Finally, we calculate the ratio $L = \frac{I_{num}}{\int_0^{\pi} e^t g(t) dt}$. When we plug our sum into the numerator, the base integral in the denominator cancels out completely!

We are left with:

$$L = \frac{e^{4\pi} - 1}{e^{\pi} - 1}$$

This result holds true for any even integer $a$. Looking at our options, both $a=2$ and $a=4$ are even integers, and they both yield this exact value for $L$.

We didn't need to perform a single complex integration. We just needed to see the structure, respect the symmetry, and let the math do the heavy lifting. That is the **JEE way**!

Animated Solution for Mathematics - Definite Integration: The option(s) with the values of $a$ and $L$ that satisfy the following equation is(are) $\frac{\int _{0}^{4 π} e ^{t} ( s i n ^{6} a t + c o s ^{4} a t ) d t}{\int _{0}^{π} e ^{t} ( s i n ^{6} a t + c o s ^{4} a t ) d t} = L ?$

Select Answer:

Visualized Solution

Define the Periodic Function $g (t)$

Periodicity of $g (t)$

Decompose the Integral

The General Interval

Apply Substitution

Simplify the General Term

Summing the Intervals

Sum of Geometric Progression

Calculate the Ratio $L$

Final Conclusion

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Mystery of the Periodic Integral

Phase 1

Phase 2

Phase 3

Phase 4

The Grand Finale

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