JEE Main 2022 (Online) 28th June Evening ShiftQuestion 1
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 $\left(\frac{\pi}{4},0\right)$, then the value of $\int_{0}^{\pi/2}y\,dx$ is equal to :
Differential Equation
Solution
Concept. $y'=2\sin x-2\tan x\,y$ is linear with IF $\sec^2x$. Formulas used. (1) $\text{IF}=e^{\int2\tan x\,dx}=\sec^2x$ and $\int2\sin x\sec^2x\,dx=2\sec x$(2) $\int_0^{\pi/2}\cos x\,dx=1$ and $\int_0^{\pi/2}\cos^2x\,dx=\frac\pi4$Step 1 - Solve. By (1), $y\sec^2x=2\sec x+C$, so $y=2\cos x+C\cos^2x$. Step 2 - Constant. Through $\left(\frac\pi4,0\right)$: $\sqrt2+\frac C2=0$, so $C=-2\sqrt2$. Step 3 - Integrate. By (2), $$\int_0^{\pi/2}y\,dx=2-2\sqrt2\cdot\frac\pi4=2-\frac{\pi}{\sqrt2}.$$ Answer: (B)