The solution of the differential equation $\cos^2 x \frac{d^2y}{dx^2} = 1$ is

The curve that satisfies the differential equation $x y \, dy - (1 + y^2) \, dx = 0$ passes through $(1, 0)$ and intersects the curve $x^2 + 3y^2 = 3$ at an angle $\theta$. Then $\frac{2\theta}{\pi} =$

Every curve represented by the general solution of $\frac{dy}{dx} = \frac{x \log x}{y^3 e^{y^2-5}}$ cuts every curve represented by the general solution of $\frac{dy}{dx} + \frac{y^3 e^{y^2-5}}{x \log x} = 0$ at an angle $\theta$. Then,$4\theta - \frac{\pi}{2} =$

If $y = \frac{x}{\ln |c x|}$ (where $c$ is an arbitrary constant) is the general solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,then the function $\phi \left( \frac{x}{y} \right)$ is:

Let $f$ be a twice differentiable function such that $f(x) = \int_{0}^{x} \tan(t-x) dt - \int_{0}^{x} f(t) \tan t dt$, where $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then $f''\left(\frac{\pi}{6}\right) + f\left(\frac{\pi}{6}\right)$ is equal to . . . . . . .

Let $f:[0, \infty) \rightarrow R$ be a continuous function such that $f(x)=1-2 x+\int_0^x e^{x-t} f(t) d t$ for all $x \in[0, \infty)$. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ The curve $y=f(x)$ passes through the point $(1,2)$
$(B)$ The curve $y=f(x)$ passes through the point $(2,-1)$
$(C)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-2}{4}$
$(D)$ The area of the region $\left\{(x, y) \in[0,1] \times R: f(x) \leq y \leq \sqrt{1-x^2}\right\}$ is $\frac{\pi-1}{4}$