"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 :"

Question

Accepted Answer

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

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

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

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

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

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

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

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

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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 :

Solution

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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 :

Solution

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