Let a curve y = y(x) be given by the solution of the differential equation cos(1/2 cos⁻¹(e⁻ˣ)) dx =

Q: Why can't I just integrate the original expression directly? It seems like a lot of work to simplify it first.

That is a very natural instinct, but in JEE Advanced, the complexity of the expression is often a signal that a direct approach will lead to an impossible integral. The original expression is a nightmare of nested functions. By simplifying first, we reduce the problem to a standard integral. The substitution $u = 1-e^{-x}$ only becomes obvious *after* you simplify the differential equation.

Q: How do I know when to use the half-angle identity? I always forget these formulas.

Whenever you see the structure $\frac{1}{2} \cos^{-1}(\dots)$, it is a massive hint. The structure $\cos(\frac{1}{2} \theta)$ is almost always solved by the half-angle identity. It is a standard pattern in JEE problems to hide a simple identity behind an inverse trigonometric function. Keep a list of these patterns; they are your best friends in the exam hall!

Q: Why did we factor out $e^{2x}$ in the denominator? That felt like a random step.

It wasn't random! We needed to transform the integral into a form where the numerator is the derivative of the function inside the square root. By factoring out $e^{2x}$, we brought an $e^{-x}$ into the numerator, which is exactly the derivative of $1-e^{-x}$. Always look for the derivative of your inner function when you see a square root!

Question

Let a curve $y = y(x)$ be given by the solution of the differential equation $\cos\left(\frac{1}{2} \cos^{-1}(e^{-x})\right) dx = \sqrt{e^{2x} - 1} dy$. If it intersects $y$-axis at $y = -1$, and the intersection point of the curve with $x$-axis is $(\alpha, 0)$, then $e^\alpha$ is equal to

Sigma Percentile · Accepted Answer

### The Art of Simplifying the Intimidating

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dissect a problem that, at first glance, looks like it was designed to make your heart race.

We have a differential equation involving nested inverse trigonometric functions and exponential terms. It is easy to feel overwhelmed, but remember: in the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

### Phase 1: The Trigonometric Veil

Look at the left-hand side of our equation: $\cos\left(\frac{1}{2} \cos^{-1}(e^{-x})ight) dx$. That nested inverse cosine is the "veil" I mentioned. It is designed to distract you.

Let us define $	heta = \cos^{-1}(e^{-x})$. By the very definition of the inverse cosine function, this implies $\cos 	heta = e^{-x}$. Now, our expression becomes $\cos(\frac{	heta}{2})$.

Do you remember your trigonometry half-angle identities? They are the keys to unlocking this door. We know that:

$$\cos\left(\frac{	heta}{2}ight) = \sqrt{\frac{1 + \cos 	heta}{2}}$$

Substituting our value for $\cos 	heta$, we get $\sqrt{\frac{1 + e^{-x}}{2}}$. By rewriting $e^{-x}$ as $\frac{1}{e^x}$, we simplify this to:

$$\sqrt{\frac{e^x + 1}{2e^x}}$$

The inverse trig function has vanished, and we are left with a clean, algebraic expression. Isn't that satisfying?

### Phase 2: The Algebraic Dance

Now, let us turn our attention to the right-hand side: $\sqrt{e^{2x} - 1} dy$. This is a classic difference of squares. We can treat $e^{2x}$ as $(e^x)^2$.

Thus, $\sqrt{e^{2x} - 1} = \sqrt{(e^x - 1)(e^x + 1)}$. When we bring both sides of our differential equation together, we get:

$$\sqrt{\frac{e^x + 1}{2e^x}} dx = \sqrt{e^x - 1} \sqrt{e^x + 1} dy$$

Look closely. Do you see it? The term $\sqrt{e^x + 1}$ appears on both sides! Since $e^x + 1$ is always positive, we can safely cancel it out.

This is the moment where the problem truly opens up. We are left with:

$$\frac{1}{\sqrt{2e^x}} dx = \sqrt{e^x - 1} dy$$

### Phase 3: The Separation of Variables

We are now in the home stretch of the calculus. We need to isolate $dy$. Dividing both sides by $\sqrt{e^x - 1}$, we get:

$$dy = \frac{1}{\sqrt{2}} \frac{dx}{\sqrt{e^x} \sqrt{e^x - 1}} = \frac{1}{\sqrt{2}} \frac{dx}{\sqrt{e^{2x} - e^x}}$$

To integrate this, we need a clever substitution. Let us factor out $e^{2x}$ from the square root: $\sqrt{e^{2x}(1 - e^{-x})} = e^x \sqrt{1 - e^{-x}}$.

Now, our equation is:

$$dy = \frac{1}{\sqrt{2}} \frac{dx}{e^x \sqrt{1 - e^{-x}}}$$

By moving $e^x$ to the numerator as $e^{-x}$, we get $dy = \frac{1}{\sqrt{2}} \frac{e^{-x} dx}{\sqrt{1 - e^{-x}}}$. This is perfectly set up for a $u$-substitution!

### Phase 4: The Final Reveal

Let $u = 1 - e^{-x}$. Then $du = e^{-x} dx$. The integral becomes:

$$\int dy = \frac{1}{\sqrt{2}} \int u^{-1/2} du$$

Integrating this, we get $y = \frac{1}{\sqrt{2}} (2u^{1/2}) + C$, which simplifies to:

$$y = \sqrt{2} \sqrt{1 - e^{-x}} + C$$

We are given that the curve passes through $(0, -1)$. Substituting $x=0$ and $y=-1$, we find $-1 = \sqrt{2} \sqrt{1 - e^0} + C$, which means $C = -1$.

Our specific curve is $y = \sqrt{2} \sqrt{1 - e^{-x}} - 1$. Finally, for the x-intercept $(\alpha, 0)$, we set $y=0$:

$$0 = \sqrt{2} \sqrt{1 - e^{-\alpha}} - 1$$

Squaring both sides gives $2(1 - e^{-\alpha}) = 1$, so $1 - e^{-\alpha} = \frac{1}{2}$, which means $e^{-\alpha} = \frac{1}{2}$.

Therefore, **$e^{\alpha} = 2$**. We have arrived at the answer. It was not about brute force; it was about seeing the patterns, respecting the identities, and trusting the process.

Enter Numerical Value:

Visualized Solution

Analyzing the Differential Equation

Simplifying the LHS: Substitution

Applying the Half-Angle Identity

Simplifying the LHS Expression

Factorizing the RHS

Combining and Simplifying the DE

Separating the Variables

Preparing for Integration

Integrating Both Sides

Applying the Initial Condition

Finding the X-intercept

Solving for $e^{α}$

The Sigma Insight: Variable Separable Method

The Art of Simplifying the Intimidating

Phase 1

Phase 2

Phase 3

Phase 4

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