Question

8 задание А вариант пожалуйста помогите

Answer 1

$intlimits^{frac{sqrt2}{2}}_0, frac{x^4dx}{sqrt{(1-x^2)^3}} , dx=Big [; x=sint; ,; dx=cost, dt,; t_1=arcsin0=0,\\ t_2=arcsinfrac{sqrt2}{2}=frac{pi}{4},; 1-x^2=1-sin^2t=cos^2t; Big ]=\\=intlimits^{frac{pi}{4}}_0; frac{sin^4tcdot cost, dt}{sqrt{cos^6t}}= intlimits^{frac{pi}{4}}_0; frac{sin^4tcdot cost, dt}{cos^3t}= intlimits^{frac{pi}{4}}_0; frac{sin^4tcdot dt}{cos^2t} =\\= intlimits^{frac{pi}{4}}_0; frac{(1-cos^2t)^2cdot dt}{cos^2t}=intlimits^{frac{pi}{4}}_0; frac{1-2cos^2t+cos^4t}{cos^2t}, dt= intlimits^{frac{pi}{4}}_0(frac{1}{cos^2t}-2+cos^2t)dt=$

$=(tgt-2t)Big |_0^{frac{pi}{4}}+intlimits^{frac{pi}{4}}_0 frac{1+cos2t}{2}dt=tgfrac{pi}{4}-2cdot frac{pi }{4}+frac{1}{2}cdot (t+frac{1}{2}sin2t)Big |_0^{frac{pi}{4}}=\\=1-frac{pi}{2}+frac{1}{2}cdot (frac{pi}{4}+frac{1}{2}cdot sinfrac{pi}{2})= 1-frac{pi}{2}+frac{pi}{8}+frac{1}{4}=frac{10-3pi }{8}$