Question

ЗНАЧЕНИЕ K=1.Помогите решить

Answer 1

$1); ; intlimits^8_1, frac{sqrt{x+2}}{x}, dx=Big [, x+2=t^2; ,; t=sqrt{x+2}; ,; x=t^2-2; ,; dx=2t, dt; Big ]=\\=Big [; t_1=sqrt{1+2}=sqrt3; ,; t_2=sqrt{8+2}=sqrt{10}; Big ]=intlimits_{sqrt3}^{sqrt{10}}, frac{tcdot 2t, dt}{t^2-2}=\\=2cdot intlimits^{sqrt{10}}_{sqrt3}, frac{t^2, dt}{t^2-2}, dt=2cdot intlimits^{sqrt{10}}_{sqrt3}, (1+frac{2}{t^2-2}), dt=2cdot (t+frac{2}{2sqrt2}cdot lnBig |frac{t-sqrt2}{t+sqrt2}Big |)Big |_{sqrt3}^{sqrt{10}}=$

$=2cdot Big (sqrt{10}-sqrt3+frac{1}{sqrt2}lnBig |frac{sqrt{10}-sqrt{3}}{sqrt{10}+sqrt3}Big |-frac{1}{sqrt2}lnBig |frac{sqrt3-sqrt2}{sqrt3+sqrt2}Big |Big )$

$2); ; intlimits^{frac{pi}{2}}_0frac{dx}{1+2cosx}=Big [, t=tgfrac{x}{2}; ,; cosx=frac{1-t^2}{1+t^2}; ,; dx=frac{2, dt}{1+t^2}; Big ]=\\=intlimits^{1}_{0}frac{2, dt}{(1+frac{2(1-t^2)}{1+t^2})cdot (1+t^2)}=intlimits^1_0frac{2, dt}{3-t^2}=frac{2}{2sqrt3}, lnBig |frac{sqrt3+t}{sqrt3-t}, Big |Big |_0^1=\\=frac{1}{sqrt3}cdot Big (lnBig |frac{sqrt3+1}{sqrt3-1}, Big |-ln|1|Big )=frac{1}{sqrt3}cdot lnBig |frac{sqrt3+1}{sqrt3-1}Big |$