Let the slope of the tangent to a curve y = f(x) at (x, y) be given by 2 [ x( x - y)].

Question

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 $$\left( {{\pi  \over 4},0} \right)$$, then the value of $$\int\limits_0^{\pi /2} {y\,dx} $$ is equal to :

Accepted Answer

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

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

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

$$\therefore$$ 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 $$\left( {{\pi  \over 4},0} \right)$$

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

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

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

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

Solutions · 1