Помогите сделать задане по мат. анализу

1) $\lim_{x \to 0} \frac{1- \sqrt{cos(x)} }{sin^2(x)} = \lim_{x \to 0} \frac{(1- \sqrt{cos(x)})(1+ \sqrt{cos(x)} ) }{(1+ \sqrt{cos(x)} )*sin^2(x)} =$
$=\lim_{x \to 0} \frac{1-cos(x)}{(1+ \sqrt{cos(x)} )*sin^2(x)} =\lim_{x \to 0} \frac{(1-cos(x))(1+cos(x))}{(1+cos(x))(1+ \sqrt{cos(x)} )*sin^2(x)} =$
$=\lim_{x \to 0} \frac{1-cos^2(x)}{(1+cos(x))(1+ \sqrt{cos(x)} )*sin^2(x)} =$
$=\lim_{x \to 0} \frac{sin^2(x)}{(1+cos(x))(1+ \sqrt{cos(x)} )*sin^2(x)} =\lim_{x \to 0} \frac{1}{(1+cos(x))(1+ \sqrt{cos(x)} )} =$
$= \frac{1}{(1+cos0)(1+ \sqrt{cos0} )} = \frac{1}{(1+1)(1+1)} = \frac{1}{4}$

2) $\int\limits^{pi}_0 {xcos \frac{x}{2} } \, dx =|u=x; dv=cos \frac{x}{2} dx;du=dx;v=2sin \frac{x}{2} |=$
$= u*v-\int\limits^{pi}_0 {v} \, du=2x*sin \frac{x}{2}- \int\limits^{pi}_0 (2sin \frac{x}{2} )dx=2x*sin \frac{x}{2}+4cos \frac{x}{2}|^{pi}_0=$
$=2pi*sin \frac{pi}{2} +4cos \frac{pi}{2} -2*0*sin0-4cos0=2pi-4$