The integrating factor of the differential equation $\cos x \frac{dy}{dx} + y \sin x = 1$ is

The integrating factor of the differential equation $(\tan ^{-1} y - x) dy = (1 + y^2) dx$ is . . . . . . .

Let $y=y(x)$ be the solution of the differential equation $\cos x(\ln(\cos x))^2 dy + (\sin x - 3y \sin x \ln(\cos x)) dx = 0$,where $x \in (0, \frac{\pi}{2})$. If $y(\frac{\pi}{4}) = \frac{-1}{\ln 2}$,then $y(\frac{\pi}{6})$ is:

The differential equation $\frac{dy}{dx} + \frac{1}{x} \sin 2y = x^3 \cos^2 y$ represents a family of curves given by:

If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-e^x \operatorname{cosec} y\right) \frac{d x}{d y}=x^5, x \geq 1$,then at $x=2$,the value of $\cos y$ is:

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x$,$x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$,such that $y(0) = 1$. Then