JEE Main 2025 — Mathematics Calculus

$$\begin{aligned} & \frac{d y}{d x}-\frac{3 \sin x}{\cos x \ln \cos x} y=-\frac{\sin x}{\cos x(\ln \cos x)^2} \\ & \text { I.F. }=e^{-\int \frac{3 \tan x}{\ln \cos x} d x} \\ & \text { Let } \ln \cos x=t \\ & -\tan x d x=d t \\ & e^{3 \int^{\frac{d t}{t}}}=e^{3 \ln t}=t^3=(\ln \cos x)^3 \\ & \therefore \quad \text { Solution will be } \\ & \quad y(\ln \cos x)^3=-\int(\tan x)(\ln \cos x) d x \\ & y(\ln \cos x)^3=\frac{(\ln \cos x)^2}{2}+c \\ & \because \quad y\left(\frac{\pi}{4}\right)=-\frac{1}{\ln 2} \\ & \Rightarrow \quad c=0 \\ & \therefore \quad y=\frac{1}{2(\ln \cos x)} \\ & y\left(\frac{\pi}{6}\right)=\frac{1}{2} \times \frac{1}{\ln \left(\cos \frac{\pi}{6}\right)} \\ & =\frac{1}{2}\left[\frac{1}{\ln \left(\frac{\sqrt{3}}{2}\right)}\right] \\ & =\frac{1}{2} \times \frac{1}{\ln \sqrt{3}-\ln 2} \\ & =\frac{1}{\ln 3-\ln 4}\end{aligned}$$​dxdy​−cosxlncosx3sinx​y=−cosx(lncosx)2sinx​ I.F. =e−∫lncosx3tanx​dx Let lncosx=t−tanxdx=dte3∫tdt​=e3lnt=t3=(lncosx)3∴ Solution will be y(lncosx)3=−∫(tanx)(lncosx)dxy(lncosx)3=2(lncosx)2​+c∵y(4π​)=−ln21​⇒c=0∴y=2(lncosx)1​y(6π​)=21​×ln(cos6π​)1​=21​​ln(23​​)1​​=21​×ln3​−ln21​=ln3−ln41​​