Match the following :
Column I
(A) ∫₀^(π/2) (sin x)^(cos x) (cos x cot x - log(sin x)^(sin x)) dx
(B

Q: Why can't I just use integration by parts for the first integral? It looks like a product of two functions.

Ah, the classic trap! While you could try, it would lead to a nightmare of algebra. In JEE Advanced, if an integral looks 'too' complex, it is almost always the derivative of a simpler function. Always check for the derivative of the base first!

Q: In the area problem, why did we integrate with respect to $y$ instead of $x$? Isn't $x$ the standard variable?

Great question. While $x$ is standard, geometry dictates the method. Since both parabolas were given as $x = f(y)$, integrating with respect to $y$ makes the limits and the subtraction trivial. Always choose the path of least resistance!

Q: How did you know to substitute $v = x+y$ in the differential equation? I would have tried separating variables directly.

You cannot separate $x$ and $y$ directly because they are locked together in the denominator. Whenever you see a function of $(ax+by+c)$ in a differential equation, that is your signal to substitute the entire expression. It turns a 'stuck' problem into a 'solved' one!

Question

Match the following :

List-I:

P. $\int_{0}^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \log(\sin x)^{\sin x}) dx$

Q. Area bounded by $-4y^2 = x$ and $x - 1 = -5y^2$

R. Cosine of the angle of intersection of curves $y = 3^{x-1} \log x$ and $y = x^x - 1$ is

S. Let $\frac{dy}{dx} = \frac{6}{x+y}$ where $y(0) = 0$ then value of $y$ when $x+y = 6$ is

List-II:

1. 1

2. 0

3. $6 \ln 2$

4. $4/3$

Sigma Percentile · Accepted Answer

### Analyzing the Hidden Derivative

Consider the integral:
$$\int_{0}^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \log(\sin x)^{\sin x}) dx$$

In competitive exams, complexity is often a mask for elegance. Whenever you see a function raised to a function, such as $(\sin x)^{\cos x}$, your first instinct should be logarithmic differentiation.

Let $f(x) = (\sin x)^{\cos x}$. Differentiating this, we obtain:
$$f'(x) = (\sin x)^{\cos x} [\cos x \cot x - \sin x \ln(\sin x)]$$

Note that $\sin x \ln(\sin x)$ is equivalent to $\ln(\sin x)^{\sin x}$. Thus, the integrand is simply the derivative of our base function. The integral evaluates to:
$$[(\sin x)^{\cos x}]_{0}^{\pi/2} = 1 - 0 = \mathbf{1}$$

### The Geometry of Parabolas

We shift to coordinate geometry with the curves $x = -4y^2$ and $x = 1 - 5y^2$. These are horizontal parabolas opening towards the negative $x$-axis.

To find the intersection points, we equate the two expressions:
$$-4y^2 = 1 - 5y^2 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$$

The area is the integral of the 'right' curve minus the 'left' curve with respect to $y$:
$$\int_{-1}^{1} ((1 - 5y^2) - (-4y^2)) dy = \int_{-1}^{1} (1 - y^2) dy$$

Evaluating this polynomial integral:
$$\left[ y - \frac{y^3}{3} \right]_{-1}^{1} = \left( 1 - \frac{1}{3} \right) - \left( -1 + \frac{1}{3} \right) = \mathbf{\frac{4}{3}}$$

### The Elegance of Tangency

We examine the intersection of $y = 3^{x-1} \log x$ and $y = x^x - 1$. We seek the cosine of the angle of intersection.

At $x=1$, both functions yield $y=0$. To find the angle, we calculate the slopes at this point:
1. For $y_1 = 3^{x-1} \log x$, the derivative at $x=1$ is $m_1 = 1$.
2. For $y_2 = x^x - 1$, the derivative at $x=1$ is $m_2 = 1$.

Since $m_1 = m_2$, the curves are tangent to each other. The angle between them is $0$, and therefore $\cos(0) = \mathbf{1}$.

### The Grand Finale: Differential Equations

Finally, we solve the differential equation:
$$\frac{dy}{dx} = \frac{6}{x+y}$$

Let $v = x+y$. Then $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1$. Substituting this into the equation:
$$\frac{dv}{dx} - 1 = \frac{6}{v} \Rightarrow \frac{dv}{dx} = \frac{v+6}{v}$$

Separating variables and integrating:
$$\int \frac{v}{v+6} dv = \int dx \Rightarrow \int \left( 1 - \frac{6}{v+6} \right) dv = x + c$$
$$v - 6 \ln|v+6| = x + c$$

Substituting $v = x+y$ back into the equation, the $x$ terms cancel, leaving $y - 6 \ln(x+y+6) = c$. Using the initial condition $y(0)=0$, we find $c = -6 \ln 6$.

For the specific case where $x+y=6$:
$$y - 6 \ln(6+6) = -6 \ln 6 \Rightarrow y = 6 \ln(12) - 6 \ln 6 = \mathbf{6 \ln 2}$$

Animated Solution for Mathematics - Differential Equations: Match the following :

List-I

List-II

Select Matching Pairs:

Visualized Solution

Part (A): Analyzing the Integrand

Part (A): The Derivative Match

Part (A): Evaluating Limits

Part (B): Visualizing the Parabolas

Part (B): Finding Intersections

Part (B): Area Integration Setup

Part (B): Final Area Calculation

Part (C): Intersection of Curves

Part (C): Slopes and Angle

Part (D): Differential Equation Setup

Part (D): Variable Separation

Part (D): Integration

Part (D): Finding the Constant

Part (D): Final Value

The Sigma Insight: Variable Separable Method

Analyzing the Hidden Derivative

The Geometry of Parabolas

The Elegance of Tangency

The Grand Finale

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