If ∫ ( 1-5 cos ^2 x sin ^5 x cos ^2 x ) d x=f(x)+ C, where C is the constant of integration, then f ( π 6 )-f ( π 4 ) 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 :

Let $f: \mathbb{R} \to \mathbb{R}$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbb{R}$ and $f'(a-1) = 0$, where $a$ is a real number. Let $g(x) = f\left( \tan^{2}x - 2\tan x + a \right), \; 0 < x < \frac{\pi}{2}$. Consider the following two statements: (I) $g$ is increasing in $\left(0, \frac{\pi}{4}\right)$ (II) $g$ is decreasing in $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then,