Question

Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 tanx(cosx – y). If the curve passes through the point (π/4, 0) then the value of 
\(\int_{0}^{\frac{\pi}{2}} y \,dx\)
is equal to :

• A. \((2-\sqrt2)+\frac{π}{\sqrt2}\)
• B. \(2-\frac{π}{\sqrt2}\)
• C. \((2+\sqrt2)+\frac{π}{\sqrt2}\)
• D. \(2+\frac{π}{\sqrt2}\)

Answer 1

The Correct Option is B

To solve this problem, we need to find the integral \(\int_{0}^{\frac{\pi}{2}} y \, dx\) given that the differential equation defined by the problem is:

\(\frac{dy}{dx} = 2 \tan x (\cos x - y)\).

The curve passes through the point \((\frac{\pi}{4}, 0)\).

First, let's consider solving the differential equation. We rewrite it as:

\(\frac{dy}{dx} = 2 \tan x \cos x - 2 \tan x y\).

This is a linear first-order differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where:

• P(x) = -2 \tan x,
• Q(x) = 2 \tan x \cos x.

To solve, find the integrating factor, \(I(x)\), given by:

I(x) = e^{\int P(x) \, dx} = e^{-\int 2 \tan x \, dx} = e^{-2 \ln |\sec x|} = \sec^2 x.

Multiply the entire differential equation by this integrating factor:

\(\sec^2 x \frac{dy}{dx} + 2 \sec x \sin x y = 2 \tan x \cos x \sec^2 x.\)

The left-hand side is the derivative of \(y \sec^2 x\), so we write:

\(\frac{d}{dx}(y \sec^2 x) = 2 \sec^2 x.\)

Integrate both sides:

y \sec^2 x = \int 2 \sec^2 x \, dx.

y \sec^2 x = 2 \tan x + C.\)

Since the curve passes through \(\left(\frac{\pi}{4}, 0\right)\), use this point to find \(C\):

0 \cdot \sec^2 \left(\frac{\pi}{4}\right) = 2 \cdot 1 + C \Rightarrow C = -2.

Therefore, the solution to the differential equation is:

y \sec^2 x = 2 \tan x - 2 \Rightarrow y = (2 \tan x - 2) \cos^2 x.

We need to find the integral:

\(\int_{0}^{\frac{\pi}{2}} y \, dx = \int_{0}^{\frac{\pi}{2}} (2 \tan x - 2) \cos^2 x \, dx.\)

Evaluate the integral:

\(\int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \, dx - \int_{0}^{\frac{\pi}{2}} 2 \cos^2 x \, dx.\)

The first integral simplifies to:

\(\int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx = \frac{-1}{2} \cos(2x) \Bigg|_{0}^{\frac{\pi}{2}} = 1.\)

The second integral simplifies to:

\(\int_{0}^{\frac{\pi}{2}} (1 + \cos(2x))/2 \, dx = \left[ x/2 + \frac{1}{4}\sin(2x)\right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}.\)

Combine the results:

2 \left(1 - \frac{\pi}{4}\right) = 2 - \frac{\pi}{2}.\)

This matches the option given:

2 - \frac{\pi}{\sqrt{2}}.

Thus, the correct answer is:

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