$$ \int \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x= $$

If $\int \frac{2 x^2+3}{\left(x^2-1\right)\left(x^2-4\right)} \mathrm{d} x=\log \left[\left(\frac{x-2}{x+2}\right)^{\mathrm{a}} \cdot\left(\frac{x+1}{x-1}\right)^{\mathrm{b}}\right]+\mathrm{c}$, (where c is the constant of integration) then the value of $a+b$ is equal to

$$ \int \frac{\sin x}{\sqrt{5 \sin ^2 x+6 \cos ^2 x}} d x $$

$$ \int \cos \left(\frac{x}{16}\right) \cdot \cos \left(\frac{x}{8}\right) \cdot \cos \left(\frac{x}{4}\right) \cdot \sin \left(\frac{x}{16}\right) \mathrm{d} x= $$