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

Answer: (B)

$2-\frac{π}{\sqrt2}$

Answer 2

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
$(\frac{π}{4},0)$
$∴c=−2\sqrt2$
$∴y=2cos⁡x−2\sqrt2cos^2⁡x$
$\therefore \int_{0}^{\frac{\pi}{2}} y \,dx = \int_{0}^{\frac{\pi}{2}} (2\cos x - 2\sqrt{2}\cos^2 x) \,dx$
$=2−2\sqrt2⋅\frac{π}{4}$
$=2−\frac{π}{\sqrt2}$