JEE Main 2022 — Mathematics Calculus

Given,

$$\frac{dy}{dx}=2\mathrm{tan}x\mathrm{cos}x-2\mathrm{tan}x\cdot y$$dxdy​=2tanxcosx−2tanx⋅y

$$\frac{dy}{dx}+(2\mathrm{tan}x)y=2\mathrm{sin}x$$dxdy​+(2tanx)y=2sinx

Now solving linear differential equation by finding Integrating factor $\displaystyle ={e}^{\int 2\mathrm{tan}xdx}=\frac{1}{{\mathrm{cos}}^{2}x}$=e∫2tanxdx=cos2x1​

Solution is given by,

$$y(\frac{1}{{\mathrm{cos}}^{2}x})=\int \frac{2\mathrm{sin}x}{{\mathrm{cos}}^{2}x}dx$$y(cos2x1​)=∫cos2x2sinx​dx

$$\Rightarrow y{\mathrm{sec}}^{2}x=\frac{2}{\mathrm{cos}x}+C$$⇒ysec2x=cosx2​+C

$$\Rightarrow y=2\mathrm{cos}x+C{\mathrm{cos}}^{2}x$$⇒y=2cosx+Ccos2x

Passes through $(\frac{\pi }{4},0)$(4π​,0)

$$\Rightarrow 0=\sqrt{2}+\frac{C}{2}\Rightarrow C=-2\sqrt{2}$$⇒0=2​+2C​⇒C=−22​

So, $f(x)=2\mathrm{cos}x-2\sqrt{2}{\mathrm{cos}}^{2}x$f(x)=2cosx−22​cos2x: Required curve

Now, $\displaystyle {\int }_{0}^{\frac{\pi }{2}}ydx=2{\int }_{0}^{\frac{\pi }{2}}\mathrm{cos}xdx-2\sqrt{2}{\int }_{0}^{\frac{\pi }{2}}{\mathrm{cos}}^{2}xdx$∫02π​​ydx=2∫02π​​cosxdx−22​∫02π​​cos2xdx

$$={[2\mathrm{sin}x]}_{0}^{\frac{\pi }{2}}-2\sqrt{2}{[\frac{x}{2}+\frac{\mathrm{sin}2x}{4}]}_{0}^{\frac{\pi }{2}}$$=[2sinx]02π​​−22​[2x​+4sin2x​]02π​​

$$=2-\frac{\pi }{\sqrt{2}}$$=2−2​π​