Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$.
• Integrating Factor (I.F.): For a first-order linear differential equation, the integrating factor is given by $I.F. = e^{\int P(x) dx}$.
• Solution of First-Order Linear Differential Equation: The solution is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Rewrite the given differential equation

We are given that the slope of the tangent is $2\tan x (\cos x - y)$. This means: $\frac{dy}{dx} = 2\tan x (\cos x - y)$ We need to rearrange this into the standard form of a first-order linear differential equation. $\frac{dy}{dx} = 2\tan x \cos x - 2\tan x \cdot y$ $\frac{dy}{dx} = 2\sin x - 2\tan x \cdot y$ $\frac{dy}{dx} + 2\tan x \cdot y = 2\sin x$ This is now in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 2\tan x$ and $Q(x) = 2\sin x$.

Step 2: Calculate the Integrating Factor (I.F.)

The integrating factor is given by: $I.F. = e^{\int P(x) \, dx} = e^{\int 2\tan x \, dx}$ $I.F. = e^{2 \int \tan x \, dx} = e^{2 \ln |\sec x|} = e^{\ln (\sec^2 x)} = \sec^2 x$ So, the integrating factor is $\sec^2 x$.

Step 3: Find the general solution

The general solution is given by: $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$ $y \cdot \sec^2 x = \int 2\sin x \cdot \sec^2 x \, dx + C$ $y \sec^2 x = 2 \int \sin x \cdot \frac{1}{\cos^2 x} \, dx + C$ Let $u = \cos x$, then $du = -\sin x \, dx$. $y \sec^2 x = 2 \int \frac{-1}{u^2} \, du + C$ $y \sec^2 x = 2 \left( \frac{1}{u} \right) + C$ $y \sec^2 x = \frac{2}{\cos x} + C$ $y \sec^2 x = 2 \sec x + C$ $y = 2\cos x + C\cos^2 x$

Step 4: Apply the initial condition to find the particular solution

We are given that the curve passes through the point $\left( \frac{\pi}{4}, 0 \right)$. Substituting these values into the general solution: $0 = 2\cos\left( \frac{\pi}{4} \right) + C\cos^2\left( \frac{\pi}{4} \right)$ $0 = 2\left( \frac{1}{\sqrt{2}} \right) + C\left( \frac{1}{\sqrt{2}} \right)^2$ $0 = \frac{2}{\sqrt{2}} + C\left( \frac{1}{2} \right)$ $-\frac{2}{\sqrt{2}} = \frac{C}{2}$ $C = -\frac{4}{\sqrt{2}} = -2\sqrt{2}$ Therefore, the particular solution is: $y = 2\cos x - 2\sqrt{2}\cos^2 x$

Step 5: Evaluate the definite integral

We need to find the value of $\int_0^{\pi/2} y \, dx$. $\int_0^{\pi/2} y \, dx = \int_0^{\pi/2} (2\cos x - 2\sqrt{2}\cos^2 x) \, dx$ $\int_0^{\pi/2} y \, dx = 2\int_0^{\pi/2} \cos x \, dx - 2\sqrt{2} \int_0^{\pi/2} \cos^2 x \, dx$ We know that $\int \cos x \, dx = \sin x$ and $\cos^2 x = \frac{1 + \cos 2x}{2}$. $\int_0^{\pi/2} y \, dx = 2[\sin x]_0^{\pi/2} - 2\sqrt{2} \int_0^{\pi/2} \frac{1 + \cos 2x}{2} \, dx$ $\int_0^{\pi/2} y \, dx = 2(\sin(\pi/2) - \sin(0)) - \sqrt{2} \int_0^{\pi/2} (1 + \cos 2x) \, dx$ $\int_0^{\pi/2} y \, dx = 2(1 - 0) - \sqrt{2} \left[ x + \frac{\sin 2x}{2} \right]_0^{\pi/2}$ $\int_0^{\pi/2} y \, dx = 2 - \sqrt{2} \left[ \left( \frac{\pi}{2} + \frac{\sin \pi}{2} \right) - \left( 0 + \frac{\sin 0}{2} \right) \right]$ $\int_0^{\pi/2} y \, dx = 2 - \sqrt{2} \left[ \frac{\pi}{2} + 0 - 0 - 0 \right]$ $\int_0^{\pi/2} y \, dx = 2 - \sqrt{2} \cdot \frac{\pi}{2} = 2 - \frac{\pi}{\sqrt{2}}$ $(2 - \sqrt 2 ) + {\pi \over {\sqrt 2 }}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs, especially when integrating and substituting. A small sign error can lead to a completely wrong answer.
• Integrating Factor: Double-check the calculation of the integrating factor. The exponent needs to be calculated correctly.
• Trigonometric Identities: Knowing trigonometric identities is crucial for simplifying the integral.

Summary

We first recognized the given equation as a first-order linear differential equation. We found the integrating factor, then the general solution, and used the initial condition to find the particular solution. Finally, we evaluated the definite integral to get the desired value. The final answer is $2 - \frac{\pi}{\sqrt{2}}$, which can be rewritten as $(2 - \sqrt{2}) + \frac{\pi}{\sqrt{2}}$.

Final Answer

The final answer is \boxed{(2 - \sqrt 2 ) + {\pi \over {\sqrt 2 }}}, which corresponds to option (A).