Question:

Rearranging the differential equation: $$\frac{dy}{dx} - \frac{3\sin x y}{\cos x \ln^2(\cos x)} = -\frac{\sin x}{\cos x(\ln\cos x)^2}$$ This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Integrating factor: $$\text{I.F.} = e^{-\int \frac{3\sin x}{\cos x \ln^2(\cos x)} dx}$$ Let $t = \ln(\cos x)$, then $dt = -\frac{\sin x}{\cos x} dx$: $$-3\int \frac{1}{t^2} dt = \frac{3}{t} = \frac{3}{\ln(\cos x)}$$ So I.F. $= e^{\frac{3}{\ln(\cos x)}} = (\cos x)^{3/\ln(\cos x)}$ Alternatively, using substitution $t = \ln(\cos x)$: $$\text{I.F.} = (\ln(\cos x))^3 = t^3$$ Multiplying the equation by I.F.: $$y(\ln(\cos x))^3 = -\int \frac{\sin x(\ln(\cos x))^2}{\cos x} dx + c$$ Using $t = \ln(\cos x)$: $$y \cdot t^3 = \int t^2 dt + c = \frac{t^3}{3} + c$$ Using initial condition $y(\pi/4) = \frac{1}{\ln 2}$: $$\frac{1}{\ln 2} \cdot (\ln(\cos(\pi/4)))^3 = \frac{(\ln(\cos(\pi/4)))^3}{3} + c$$ Since $\ln(\cos(\pi/4)) = \ln(1/\sqrt{2}) = -\frac{\ln 2}{2}$, we get $c = 0$. Therefore: $$y = \frac{1}{2(\ln(\cos x))}$$ At $x = \pi/6$: $$y(\pi/6) = \frac{1}{2\ln(\cos(\pi/6))} = \frac{1}{2\ln(\sqrt{3}/2)} = \frac{1}{2(\ln\sqrt{3} - \ln 2)} = \frac{1}{\ln 3 - \ln 4} = \log_e 3 - \log_e 4$$