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

The correct answer is (B) : \(2-\frac{π}{\sqrt2}\)
\(\frac{dy}{dx}=2tan⁡x(cos⁡x−y)\)
\(⇒\frac{dy}{dx}+2tan⁡xy=2sin⁡x\)
\(I.F=e^{∫2tan⁡xdx}=sec^2⁡x\)
∴ Solution of D.E. will be
\(y(x)sec^2⁡x=∫2sin⁡xsec^2⁡xdx\)
\(ysec^2⁡x=2sec⁡x+c\)
∵ Curve passes through
\((\frac{π}{4},0)\)
\(∴c=−2\sqrt2\)
\(∴y=2cos⁡x−2\sqrt2cos^2⁡x\)
\(\therefore \int_{0}^{\frac{\pi}{2}} y \,dx = \int_{0}^{\frac{\pi}{2}} (2\cos x - 2\sqrt{2}\cos^2 x) \,dx\)
\(=2−2\sqrt2⋅\frac{π}{4}\)
\(=2−\frac{π}{\sqrt2}\)