Question

помогите пожалуйста)) Тема: Интегрирование по частям
хоть один плз

Answer 1

$5); ; intlimits^{frac{pi}{2}}_0sinxcdot cos^2x, dx=[; t=cosx; ,; dt=-sinx, dx; ,; t_1=cos0=1; ,\\t_2=cosfrac{pi}{2}=0]=-intlimits^0_1t^2, dt=-frac{t^3}{3}|_1^0=-frac{1}{3}(0^3-1^3)=frac{1}{3}; ;\\6); ; intlimits^{sqrt{a}}_0 {x^2sqrt{a-x^2}} , dx =[; x=sqrt{a}cdot sint; ,; dx=sqrt{a}cdot cost, dt; ,\\t=arcsinfrac{x}{sqrt{a}}; ,; t_1=0; ,t_2=arcsin1=frac{pi}{2}; ]=$

$=intlimits_0^{frac{pi}{2}} {acdot sin^2tsqrt{a-asin^2t}} cdot sqrt{a}cdot cost, dt=intlimits^{frac{pi}{2}}_0, asin^2tcdot sqrt{acdot cos^2t}cdot sqrt{a}cdot cost, dt=$

$=a^2 intlimits^{frac{pi}{2}}_0sin^2tcdot cos^2t, dt=\\=[; sintcdot cost=frac{1}{2}sin2t; ]=a^2 intlimits^{frac{pi}{2}}_0; frac{1}{4}sin^22t; dt=\\=frac{a^2}{4} intlimits^{frac{pi}{2}}_0 frac{1-cos4t}{2} dt= frac{a^2}{8} intlimits^{frac{pi}{2}}_0(1-cos4t)dt= frac{a^2}{8}(t-frac{1}{4}sin4t)|_0^{frac{pi}{2}}=\\= frac{a^2}{8}(frac{pi}{2}- frac{1}{4}sinpi )=frac{a^2}{8}(frac{pi}{2}-0)=frac{pi a^2}{16}$

$8); ; int _0^1sqrt{1+x^2}dx; ;$

$A=int sqrt{1+x^2}dx=int frac{1+x^2}{sqrt{1+x^2}}dx=int frac{dx}{sqrt{1+x^2}} +int frac{xcdot x, dx}{sqrt{1+x^2}} =\\=ln|x+sqrt{1+x^2}|+[, u=x; ,du=dx; ,; dv=frac{x, dx}{sqrt{1+x^2}}; ,\\v=frac{1}{2}int frac{2x, dx}{sqrt{1+x^2}}=frac{1}{2}int frac{dz}{sqrt{z}} = frac{1}{2} cdot 2sqrt{z}=sqrt{z}=sqrt{1+x^2}; ]=\\=ln|x+sqrt{1+x^2}|+xcdot sqrt{1+x^2}underbrace{-int sqrt{1+x^2}dx}_{-A}; ;\\A=ln|x+sqrt{1+x^2}|+xcdot sqrt{1+x^2}-A; ;\\2A=2cdot int sqrt{1+x^2}dx=ln|x+sqrt{1+x^2}|+xcdot sqrt{1+x^2}; ;$

$A=int sqrt{1+x^2}dx=frac{1}{2}cdot Big (ln|x+sqrt{1+x^2}|+xcdot sqrt{1+x^2}Big )\\\int limits _0^1sqrt{1+x^2}dx=frac{1}{2}cdot Big (ln|x+sqrt{1+x^2}|+xcdot sqrt{1+x^2}Big)Big |_0^1=\\=frac{1}{2}cdot Big (ln|1+sqrt2|+1cdot sqrt2-ln|0+1|-0Big )=\\=frac{1}{2}cdot Big (sqrt2+ln(1+sqrt2)Big )=frac{sqrt2+ln(1+sqrt2)}{2}; ;$