Question

Определённые интегралы

Answer 1

$1); ; intlimits^1_0, frac{x, dx}{x^4+3}= intlimits^1_0,frac{x, dx}{(x^2)^2+3} =[, t=x^2; ,; dt=2x, dx; ,; t_1=0; ,; t_2=1]=\\= frac{1}{2}, intlimits^1_0, frac{dt}{t^2+3}= frac{1}{2sqrt3}, arctg frac{t}{sqrt3}; Big |_0^1=frac{1}{2sqrt3}, arctg frac{x^2}{sqrt3}; Big |_0^1=\\= frac{1}{2sqrt3} , (arctg frac{1}{sqrt3}-0)= frac{1}{2sqrt3}cdot frac{pi }{6} = frac{pi }{12sqrt3}$

$2); ; intlimits^1_0 , frac{x^4dx}{(2-x^2)^{3/2}}=[, x=sqrt2sint; ,dx=sqrt2cost, dt; ,; t=arcsinfrac{x}{sqrt2}; ,\\t_1=0; ,; t_2= frac{pi }{4}; ]= intlimits^{frac{pi}{4}}_0 , frac{4sin^4tcdot sqrt2cost, dt}{sqrt{(2-2sin^2t)^3}}= intlimits^{ frac{pi}{4}}_0, frac{4sqrt2, sin^4t, cost, dt}{sqrt{(2cos^2t)^3}} =\\= intlimits^{frac{pi}{4}}_0, frac{4sqrt2, sin^4t, cost, dt}{2sqrt2, cos^3t} = 2, intlimits^{ frac{pi}{4}}_0, frac{(sin^2t)^2}{cos^2t}, dt=2, intlimits^{frac{pi}{4}}_0, frac{(1-cos^2t)^2}{cos^2t}, dt=$

$=2, intlimits^{frac{pi}{4}}_0, frac{1-2cos^2t+cos^4t}{cos^2t}, dt=2 intlimits^{frac{pi}{4}}_0( frac{1}{cos^2t}-2+cos^2t)dt=\\=2(tgt-2t)Big |_0^{frac{pi}{4}}+ intlimits^{frac{pi}{4}}_0, (1+cos2t)dt=\\=2(tgfrac{pi}{4}-2cdot frac{pi}{4})+(t+frac{1}{2}, sin2t)Big |_0^{frac{pi }{4}}=2(1- frac{pi}{2})+frac{pi}{4}+frac{1}{2}=\\=2-pi + frac{pi }{4}+frac{1}{2}=- frac{3pi }{4}+ frac{5}{2}$

$3); ; intlimits^1_0, (3x-2), 3^{1-x}, dx=[, u=3x-2; ,; du=3, dx; ,; dv=3^{1-x}dx, \\v=- frac{3^{1-x}}{ln3}; ]=(3x-2)cdot frac{3^{1-x}}{ln3}; Big |_0^1+frac{3}{ln3}, intlimits^1_0, 3^{1-x}, dx=\\=frac{1}{ln3}- frac{2cdot 3}{ln3}+ frac{3}{ln3}cdot frac{-3^{1-x}}{ln3}, Big |_0^1=- frac{5}{ln3}- frac{3}{ln^23}cdot (3^0-3)=-frac{5}{ln3}+frac{6}{ln^23}$