The function f:R ~0 → R given by f(x)= 1 x- 2 e^2 x-1 can be made continuous at x=0 by defining f(0) as

If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to

If the solution curve $y=f(x)$ of the differential equation $\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.

Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi}\left(\frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .

Let $P_{1}: y=4 x^{2}$ and $P_{2}: y=x^{2}+27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y=\alpha x, \alpha>0$ and $P_{1}$, then $\alpha$ is equal to :

If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :