QJEEM22Ma4617: JEE Main 2022 Mathematics - Class 12 JEEM22Ma4617 Solution: Let the slope of the tangent to a curve y = f(x) at (x, y) be…

${{dy} \over {dx}} = 2\tan x(\cos x - y)$

$\Rightarrow {{dy} \over {dx}} + 2\tan xy = 2\sin x$

$I.F. = {e^{\int {2\tan xdx} }} = {\sec ^2}x$

$\therefore$ Solution of D.E. will be

$y(x){\sec ^2}x = \int {2\sin x{{\sec }^2}xdx}$

$y{\sec ^2}x = 2\sec x + c$

$\because$ Curve passes through $\left( {{\pi \over 4},0} \right)$

$\therefore$ $c = - 2\sqrt 2$

$\therefore$ $y = 2\cos x - 2\sqrt 2 {\cos ^2}x$

$\therefore$ $\int_0^{\pi /2} {ydx = \int_0^{\pi /2} {(2\cos x - 2\sqrt 2 {{\cos }^2}x)\,dx} }$

$= 2 - 2\sqrt 2 \,.\,{\pi \over 4} = 2 - {\pi \over {\sqrt 2 }}$