Question

Срочно. Даю 75 баллов

Answer 1

Интегрирование по частям.$\int udv=uv-\int vdu\\\int x^2\cos xdx=\left(\begin{array}{cc}u=x^2&du=2xdx\\dv=\cos xdx&v=\sin x\end{array}\right)=\\=x^2\sin x-2\int x\sin xdx=\left(\begin{array}{cc}u=x&du=dx\\dv=\sin xdx&v=-\cos x\end{array}\right)=\\=x^2\sin x+2x\cos x-2\int\cos xdx=x^2\sin x+2x\cos x-2\sin x+C=\\=(x^2-2)\sin x+2x\cos x+C$

$\int\limits_b^a\ln xdx=\left.(x\ln x-x)\right|\limits_b^a=\left.(x(\ln x-1))\right|\limits_b^a=a(\ln a-1)-b(\ln b-1)\\\\\int\limits_b^ax^2\cdot l^xdx=\left(\begin{array}{cc}u=x^2&du=2xdx\\dv=l^xdx&v=\frac l{\ln l}\end{array}\right)=\left.\frac{x^2l^x}{\ln l}\right|\limits_b^a-\frac2{\ln l}\int\limits_b^ax\cdotl^xdv=\\=\frac{a^2l^a-b^2l^b}{\ln l}-\frac2{\ln l}\int\limits_b^ax\cdotl^xdv=\left(\begin{array}{cc}u=x&du=dx\\dv=l^xdx&v=\frac l{\ln l}\end{array}\right)=$
$=\frac{a^2l^a-b^2l^b}{\ln l}+\left.\left(-\frac{2xl^x}{\ln^2l}\right)\right|\limits_b^a+\frac2{\ln^2l}\int\limits_b^al^xdx=\frac{a^2l^a-b^2l^b}{\ln l}-\frac{2al^a-2bl^b}{\log^2l}+\left.\frac{2l^x}{\ln^3l}\right|\limits_b^a=\\=\frac{a^2l^a-b^2l^b}{\ln l}-\frac{2al^a-2bl^b}{\log^2l}+\frac{2l^a-2l^b}{\ln^3l}$

$\int\limits_b^ax\cos xdx=\left(\begin{array}{cc}u=u&du=dx\\dv=\cos xdx&v=\sin x\end{array}\right)=\left.x\sin x\right|\limits_b^a-\int\limits_b^a\sin xdx=\\=a\sin a-b\sin b+\left.\cos x\right|\limits_b^a=a\sin a-b\sin b+\cos a-\sin b$