Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tan x (cos x - y). If the c

Q: Why can't we just use variable separation here? It looks like a simple equation.

That is a classic trap! Variable separation works when you can write the equation as $f(y) dy = g(x) dx$. Here, the term $-y$ is trapped inside the parenthesis with $\cos x$. Because of that addition/subtraction, you cannot isolate $y$ and $x$ completely. That is exactly why we must pivot to the Linear Differential Equation method.

Q: Is there a trick to remembering the Integrating Factor formula?

Think of it as the 'reverse product rule.' We are essentially trying to force the left side of the equation to look like the derivative of a product, $\frac{d}{dx}(y \cdot \text{I.F.})$. The formula $e^{\int P dx}$ is the mathematical 'glue' that makes that product rule possible. Don't just memorize it; visualize it as the tool that turns a messy sum into a clean derivative.

Q: Why did we use Wallis' formula for the final integral? Can't we just use the half-angle identity?

You absolutely can use the half-angle identity $\cos^2 x = \frac{1 + \cos 2x}{2}$! It is perfectly valid. However, in the heat of a JEE Advanced exam, time is your most precious currency. Wallis' formula is a powerful shortcut for integrals of the form $\int_0^{\pi/2} \sin^n x dx$ or $\int_0^{\pi/2} \cos^n x dx$. Mastering it gives you an edge.

Question

Let the slope of the tangent to a curve $y = f(x)$ at (x, y) be given by $2 	an x (\cos x - y)$. If the curve passes through the point $(\pi/4, 0)$, then the value of $\int_0^{\pi/2} y dx$ is equal to

Options:

(A) $(2 - \sqrt{2}) + \frac{\pi}{\sqrt{2}}$

(B) $2 - \frac{\pi}{\sqrt{2}}$

(C) $(2 + \sqrt{2}) + \frac{\pi}{\sqrt{2}}$

(D) $2 + \frac{\pi}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### The Art of Decoding Change

Welcome, future engineers. Today, we are not just solving a differential equation; we are uncovering the hidden geometry of a curve.

When you see a problem like this, where the slope of a tangent is defined by a relationship between $x$ and $y$, do not panic. You are looking at a dynamic system.

The slope, $\frac{dy}{dx}$, is the language of change. Our goal is to translate this language into a static equation of a curve, $y = f(x)$, and then calculate the area beneath it.

### Phase 1: Identifying the Hidden Structure

We start with the given expression:

$$ \frac{dy}{dx} = 2 	an x (\cos x - y) $$

If we distribute the $2 	an x$, we get:

$$ \frac{dy}{dx} = 2 	an x \cos x - 2 y 	an x $$

Since $	an x = \frac{\sin x}{\cos x}$, the first term simplifies beautifully: $2 	an x \cos x = 2 \sin x$.

Now, move the $y$ term to the left side:

$$ \frac{dy}{dx} + (2 	an x) y = 2 \sin x $$

Suddenly, the fog clears. This is the classic standard form of a Linear Differential Equation: $\frac{dy}{dx} + P(x)y = Q(x)$. Identifying this form is the moment you take control of the problem.

### Phase 2: The Magic of the Integrating Factor

In a linear differential equation, the Integrating Factor (I.F.) is our secret weapon. It acts as a multiplier that transforms the left side of our equation into the derivative of a product.

We calculate it as $I.F. = e^{\int P(x) dx}$. Here, $P(x) = 2 	an x$.

Integrating this, we get:

$$ \int 2 	an x dx = 2 \ln |\sec x| $$

When we raise $e$ to this power, $e^{2 \ln |\sec x|}$, the logarithm and the exponential function cancel out, leaving us with $\sec^2 x$. This is the elegant 'glue' that will hold our solution together.

### Phase 3: The General Solution

With our I.F. in hand, the general solution is simply $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$. Substituting our values, we get:

$$ y \sec^2 x = \int (2 \sin x) \cdot \sec^2 x dx $$

Do not let this integral intimidate you. Rewrite $\sec^2 x$ as $\frac{1}{\cos^2 x}$.

Then, the integrand becomes:

$$ 2 \sin x \cdot \frac{1}{\cos^2 x} = 2 \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = 2 	an x \sec x $$

We know the integral of $\sec x 	an x$ is simply $\sec x$. Thus, our equation becomes $y \sec^2 x = 2 \sec x + C$. We have found the family of curves that satisfy the differential condition.

### Phase 4: Finding the Specific Path

We are not done yet. The problem gives us a specific point: $(\frac{\pi}{4}, 0)$. This is our boundary condition.

By substituting $x = \frac{\pi}{4}$ and $y = 0$, we find:

$$ 0 = 2 \sec\left(\frac{\pi}{4}ight) + C $$

Since $\sec(\frac{\pi}{4}) = \sqrt{2}$, we get $0 = 2\sqrt{2} + C$, which means $C = -2\sqrt{2}$.

Now, we have the specific equation of our curve: $y \sec^2 x = 2 \sec x - 2\sqrt{2}$. Dividing by $\sec^2 x$ (or multiplying by $\cos^2 x$), we get the explicit form:

$$ y = 2 \cos x - 2\sqrt{2} \cos^2 x $$

### Phase 5: The Final Integration

Finally, we calculate the area:

$$ I = \int_0^{\pi/2} (2 \cos x - 2\sqrt{2} \cos^2 x) dx $$

We split this into two parts. The first, $\int_0^{\pi/2} 2 \cos x dx$, is straightforward: $2 [\sin x]_0^{\pi/2} = 2(1 - 0) = 2$.

The second part, $-2\sqrt{2} \int_0^{\pi/2} \cos^2 x dx$, is where we use Wallis' formula. The integral of $\cos^2 x$ from $0$ to $\frac{\pi}{2}$ is $\frac{\pi}{4}$.

Thus, the second term is:

$$ -2\sqrt{2} \cdot \frac{\pi}{4} = -\frac{\pi}{\sqrt{2}} $$

Combining them, we arrive at our final answer: **$2 - \frac{\pi}{\sqrt{2}}$**. You have successfully navigated the entire process, from a differential equation to a definite integral. This is the power of calculus—modeling the world, one step at a time.

Animated Solution for Mathematics - Differential Equations: Let the slope of the tangent to a curve $y = f (x)$ at (x, y) be given by $2 tan x (cos x - y)$ . If the curve passes through the point $(π /4, 0)$ , then the value of $\int_{0}^{π /2} y d x$ is equal to

Select Answer:

Visualized Solution

Slope of Tangent

Rearranging the Equation

Linear Differential Equation

Integrating Factor (I.F.)

General Solution Setup

Solving the Integral

Applying Initial Condition

Finding the Constant C

Equation of the Curve

Setting up the Definite Integral

Integrating the First Part

Integrating the Second Part

Final Result

The Sigma Insight: Linear Differential Equations

The Art of Decoding Change

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})$ . If $lim_{x \to 0^{+}} x y (x) = 1$ , then the value of $y (\frac{π}{4})$ is :

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} = (tan x - y) sec^{2} x, x \in (- π /2, π /2)$ , such that $y (0) = 0$ , then $y (- π /4)$ is equal to :

The function $y = f (x)$ is the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{4} + 2 x}{1 - x ^{2}}$ in $(- 1, 1)$ satisfying $f (0) = 0$ . Then $\int_{- 3 /2}^{3 /2} f (x) d x$ is

If $y = y (x)$ is the solution of the differential equation $\frac{d y}{d x} + (tan x) y = sin x, 0 \leq x \leq \frac{π}{3}$ , with $y (0) = 0$ , then $y (\frac{π}{4})$ equal to :

Let $y = y (x)$ be the solution of the differential equation, $\frac{d y}{d x} + y tan x = 2 x + x^{2} tan x$ , $x \in (- π /2, π /2)$ , such that $y (0) = 1$ . Then :

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0, y (0) = 1$ . Then the value of $y (1)$ is :

Let $y = y (x)$ be the solution of the differential equation $cos x \frac{d y}{d x} + 2 y sin x = sin 2 x$ , $x \in (0, \frac{π}{2})$ . If $y (π /3) = 0$ , then $y (π /4)$ is equal to :

Let $y = y (x)$ be the solution curve of the differential equation $sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0, 0 < x < \frac{π}{2}$ , which passes through the point $(\frac{π}{6}, 1)$ . Then $∣ y (\frac{π}{3}) ∣$ is equal to

If $y (x)$ satisfies the differential equation $y^{'} - y tan x = 2 x sec x$ and $y (0) = 0$ , then

Let $y = y (x)$ be the solution of the differential equation $x d y = (y + x^{3} cos x) d x$ with $y (π) = 0$ , then $y (\frac{π}{2})$ is equal to: