Question

Решите пожалуйста примеры.
UPD: в 4 примере если не видно 2-x^2 а внизу 2x-1.

Answer 1

$U=z*tg(3x-2y)\frac{delta U}{delta x}=z*frac{1}{cos^2(3x-2y)}*3=frac{3z}{cos^2(3x-2y)}\frac{delta U}{delta y}=z*frac{1}{cos^2(3x-2y)}*(-2)=frac{-2z}{cos^2(3x-2y)}\frac{delta U}{delta z}=tg(3x-2y)$


$intlimits^1_{-3}dx intlimits^{2-x^2}_{2x-1}(x-y)dy= intlimits^1_{-3}(x(2-x^2)-frac{(2-x^2)^2}{2}-x(2x-1)+\+frac{(2x-1)^2}{2})dx=...\\intlimits^{2-x^2}_{2x-1}(x-y)dy=xy-frac{y^2}{2}|^{2-x^2}_{2x-1}=x(2-x^2)-frac{(2-x^2)^2}{2}-x(2x-1)+\+frac{(2x-1)^2}{2}=x(2-x^2-2x+1)-frac{1}{2}((2-x^2)^2-(2x-1)^2)=\=x(-x^2-2x+3)-frac{1}{2}((2-x^2-2x+1)(2-x^2+2x-1))=\=(-x^2-2x+3)(x-frac{1}{2}(-x^2+2x+1))=\=(-x^2-2x+3)(x+frac{x^2}{2}-x-frac{1}{2})=(-x^2-2x+3)(frac{x^2-1}{2})=$
$=frac{1}{2}(-x^4-2x^3+4x^2+2x-3)\...=frac{1}{2}intlimits^1_{-3}(-x^4-2x^3+4x^2+2x-3)dx=\=frac{1}{2}(-frac{x^5}{5}-frac{x^4}{2}+frac{4x^3}{3}+x^2-3x)|^1_{-3}=\\=frac{1}{2}(-frac{1}{5}-frac{1}{2}+frac{4}{3}+1-3-frac{243}{5}+frac{81}{2}+36-9-9)=\\=frac{1}{2}(-frac{244}{5}+47+frac{4}{3})=frac{1}{2}(-48frac{4}{5}+56+1frac{1}{3})=frac{1}{2}(9-frac{7}{15})=\\=frac{1}{2}(8frac{8}{15})=frac{1}{2}*frac{128}{15}=frac{64}{15}=4frac{4}{15}$


$(1+x^2)y'+y=arctgx\y=uv;y'=u'v+v'u\(1+x^2)u'v+u((1+x^2)v'+v)=arctgx\begin{cases}(1+x^2)v'+v=0\(1+x^2)u'v=arctgxend{cases}\(1+x^2)frac{dv}{dx}+v=0|*-frac{1}{dv}\-frac{1+x^2}{dx}=frac{v}{dv}\-frac{dx}{1+x^2}=frac{dv}{v}\-intfrac{dx}{1+x^2}=intfrac{dv}{v}\-arctgx=ln|v|\v=e^{-arctgx}\frac{(1+x^2)e^{-arctgx}du}{dx}=arctgx|*frac{dx}{(1+x^2)e^{-arctgx}}\du=frac{e^{arctgx}*arctgxdx}{1+x^2}\int du=intfrac{e^{arctgx}*arctgxdx}{1+x^2}$
$int e^{arctgx}*arctgx d(arctgx)=arctgx*e^{arctgx}}-int e^{arctgx}d(arctgx)=\=e^{arctgx}(arctgx-1)+C\u=arctgx= textgreater du=d(arctgx)\dv=e^{arctgx}d(arctgx)= textgreater v=e^{arctgx}\\u=e^{arctgx}(arctgx-1)+C\y=(e^{arctgx}(arctgx-1)+C)e^{-arctgx}=arctgx-1+frac{C}{e^{arctgx}}$