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
- A \((2-\sqrt{2})+\frac{\pi}{\sqrt{2}}\)
- B \(2-\frac{\pi}{\sqrt{2}}\)
- C \((2+\sqrt{2})+\frac{\pi}{\sqrt{2}}\)
- D \(2+\frac{\pi}{\sqrt{2}}\)
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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