Question

Помогите сделать задане по мат. анализу

Answer 1

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