If I = (2/π) ∫_{-π/4}^{π/4} dx / ((1 + e^{sin x})(2 - cos 2x)), then 27I² equals

Q: Wait, why does the term $\frac{1}{1 + e^{\sin x}} + \frac{1}{1 + e^{-\sin x}}$ simplify to 1? Is that always true?

It is not always true, but it is a beautiful property of functions of the form $f(x) = \frac{1}{1 + a^x}$. When you add $f(x) + f(-x)$, the $a^x$ terms cancel out perfectly. In our case, $a = e$, so it works like magic! Always look for this symmetry when you see exponential functions in the denominator.

Q: Why did we use $\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x}$ instead of just $\cos 2x = 2\cos^2 x - 1$?

Great question! While both are correct, the $\tan x$ identity is a 'power move' here because its derivative, $\sec^2 x$, is exactly what we need to perform a substitution. If you used the cosine identity, you would end up with $\sec^2 x$ in the denominator, which is much harder to integrate.

Q: Why can we change the limits from $[-\pi/4, \pi/4]$ to $[0, \pi/4]$ and multiply by 2?

We can do this because the function $f(x) = \frac{1}{2 - \cos 2x}$ is an even function. Since $\cos 2(-x) = \cos 2x$, the function is symmetric across the y-axis. Integrating from $-a$ to $a$ is the same as doubling the area from $0$ to $a$.

Q: I got stuck at the substitution $t = \sqrt{3} \tan x$. How do I know what constant to pick?

Look at the denominator after simplification: $1 + 3\tan^2 x$. We want this to look like $1 + t^2$. By setting $t^2 = 3\tan^2 x$, we immediately see that $t = \sqrt{3} \tan x$. Always match your substitution to the structure of the denominator!

Q: Is there any other way to solve $\int \frac{dx}{2 - \cos 2x}$ without using $\tan x$?

You could use the Weierstrass substitution ($t = \tan x$), but that is essentially what we did! You could also try complex numbers or contour integration, but for JEE Advanced, the trigonometric substitution is the most elegant and time-efficient path.

Question

If $I = \frac{2}{\pi} \int_{-\pi/4}^{\pi/4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$, then $27I^2$ equals

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we stand before an integral that looks, at first glance, like a chaotic mess of exponentials and trigonometric functions.

You might feel a slight tremor of hesitation—that is perfectly normal. But remember, in the world of JEE Advanced, complexity is often just a mask for hidden elegance. Let us peel back that mask together.

We begin with our integral:

$$I = \frac{2}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$$

### The Power of King's Rule

Whenever you see symmetric limits like $[-\frac{\pi}{4}, \frac{\pi}{4}]$, your intuition should immediately scream 'King's Rule!' This property, $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, is the ultimate tool for simplifying symmetric integrands.

Since our sum of limits $a+b$ is $0$, replacing $x$ with $-x$ is our first move. As we substitute, we use the fact that $\sin(-x) = -\sin x$ and $\cos(-2x) = \cos 2x$.

The integral transforms into:

$$I = \frac{2}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{(1 + e^{-\sin x})(2 - \cos 2x)}$$

### The Beautiful Cancellation

Now, watch the magic happen. When we add our original integral to this new version, we get $2I$. Look at the bracketed term:

$$\frac{1}{1 + e^{\sin x}} + \frac{1}{1 + e^{-\sin x}}$$

If you multiply the second fraction by $\frac{e^{\sin x}}{e^{\sin x}}$, it becomes $\frac{e^{\sin x}}{e^{\sin x} + 1}$. When you add this to the first term, the numerators and denominators match perfectly!

The entire exponential mess collapses into $1$. We are left with the much cleaner:

$$2I = \frac{2}{\pi} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{dx}{2 - \cos 2x}$$

### Exploiting Even Symmetry

Since $f(x) = \frac{1}{2 - \cos 2x}$ is an even function, we can simplify our work by integrating from $0$ to $\frac{\pi}{4}$ and doubling the result. This gives us:

$$I = \frac{2}{\pi} \int_0^{\frac{\pi}{4}} \frac{dx}{2 - \cos 2x}$$

Now, how do we handle the trigonometric denominator? We use the half-angle identity:

$$\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$$

Substituting this into our integral, we get:

$$I = \frac{2}{\pi} \int_0^{\frac{\pi}{4}} \frac{dx}{2 - \frac{1 - 	an^2 x}{1 + 	an^2 x}}$$

### The Final Transformation

After simplifying the complex fraction, the $1 + 	an^2 x$ term jumps to the numerator, becoming $\sec^2 x$. Our integral now looks like this:

$$I = \frac{2}{\pi} \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{1 + 3	an^2 x} dx$$

This is the moment of truth. We set $t = \sqrt{3} 	an x$, which implies $dt = \sqrt{3} \sec^2 x dx$. Our limits change from $[0, \frac{\pi}{4}]$ to $[0, \sqrt{3}]$.

The integral becomes:

$$I = \frac{2}{\pi\sqrt{3}} \int_0^{\sqrt{3}} \frac{dt}{1 + t^2}$$

### The Grand Finale

This is a standard form! The integral of $\frac{1}{1+t^2}$ is simply $	an^{-1} t$. Evaluating this from $0$ to $\sqrt{3}$, we get $	an^{-1}(\sqrt{3}) - 	an^{-1}(0) = \frac{\pi}{3}$.

Finally, we calculate $I$:

$$I = \frac{2}{\pi\sqrt{3}} 	imes \frac{\pi}{3} = \frac{2}{3\sqrt{3}}$$

Squaring this gives $I^2 = \frac{4}{27}$. Therefore, **$27I^2 = 27 	imes \frac{4}{27} = 4$**.

Take a deep breath. You have navigated through symmetry, trigonometric identities, and substitution to reach the summit. The final answer is **$4$**.

Animated Solution for Mathematics - Definite Integration: If $I = \frac{2}{π} \int_{- π /4}^{π /4} \frac{d x}{( 1 + e ^{s i n x} ) ( 2 - c o s 2 x )}$ , then $27 I^{2}$ equals

Enter Numerical Value:

Visualized Solution

Original Integral & King's Rule

Applying King's Rule

Adding the Integrals

Simplifying the Bracket

Even Function Property

Half-Angle Identity

Simplifying the Denominator

Substitution Method

Transformed Integral

Evaluating the Integral

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

The Power of King's Rule

The Beautiful Cancellation

Exploiting Even Symmetry

The Final Transformation

The Grand Finale

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If I=π2​∫−π/4π/4​(1+esinx)(2−cos2x)dx​, then 27I2 equals