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JEE Mains · Maths · STD 12 - 9. 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 \((\frac{\pi}{4},0)\), then the value of \(\int \limits_{0}^{\pi / 2} y d x\) is equal to

  1. A \((2-\sqrt{2})+\frac{\pi}{\sqrt{2}}\)
  2. B \(2-\frac{\pi}{\sqrt{2}}\)
  3. C \((2+\sqrt{2})+\frac{\pi}{\sqrt{2}}\)
  4. D \(2+\frac{\pi}{\sqrt{2}}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

\(\frac{d y}{d x}=2 \tan x \cos x-2 \tan x \cdot y\) \(\frac{d y}{d x}+(2 \tan x) y=2 \sin x\) Integrating factor \(=e^{\int 2 \tan x d x}=\frac{1}{\cos ^{2} x}\) \(y\left(\frac{1}{\cos ^{2} x}\right)=\int \frac{2 \sin x}{\cos ^{2} x} d x\) \(y \sec ^{2} x=\frac{2}{\cos x}+C\)…
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