35 БАЛЛОВ ЗА 2 ИНТЕГРАЛА ПОМОГИТЕ

(11)
$\int\limits^{e}_{1} {x^2*ln(x)} \, dx =\frac{1}{3}* \int\limits^{e}_{1} {ln(x)} \, d(x^3) =\\\\=\frac{1}{3}*[ x^3*ln(x)|^e_1-\int\limits^{e}_{1} {x^3*[ln(x)]'} \, dx] =\\\\ =\frac{1}{3}*[ e^3*ln(e)-1*ln(1)-\int\limits^{e}_{1} {x^3*\frac{1}{x} \, dx] =\\\\$

$=\frac{1}{3}*[ e^3-\int\limits^{e}_{1} {x^2 \, dx] =\frac{1}{3}*[ e^3-\frac{x^3|^e_1}{3}] =\frac{1}{3}*[ e^3-\frac{e^3-1}{3}] =\frac{2e^3+1}{9}$

(12)
$I= \int\limits {e^x*cos(x)} \, dx =\int\limits {cos(x)} \, d(e^x) =\\\\ =e^x*cos(x)- \int\limits {e^x} \, d[cos(x)] =\\\\ =e^x*cos(x)- \int\limits {e^x*[-sin(x)]} \, dx =\\\\ =e^x*cos(x)+ \int\limits {e^x*sin(x)} \, dx =\\\\ =e^x*cos(x)+ \int\limits {sin(x)} \, d(e^x) =\\\\ =e^x*cos(x)+e^x*sin(x)- \int\limits {e^x} \, d[sin(x)] =\\\\ =e^x*cos(x)+e^x*sin(x)- \int\limits {e^x*cos(x)} \, dx =\\\\ =e^x*cos(x)+e^x*sin(x)- I .\\\\ 2I=e^x*cos(x)+e^x*sin(x)\\\\$

$I=\frac{e^x*cos(x)+e^x*sin(x)}{2}=\frac{e^x}{2}*[sin(x)+cos(x)]\\\\ -----------------------------\\ \frac{e^x}{2}*[sin(x)+cos(x)]\ |^{\frac{\pi}{2}}_0=\\\\ =\frac{e^{\frac{\pi}{2}}}{2}*[sin(\frac{\pi}{2})+cos(\frac{\pi}{2})]-\frac{e^0}{2}*[sin(0)+cos(0)]=\\\\ =\frac{e^{\frac{\pi}{2}}}{2}*[1+0]-\frac{1}{2}*[0+1]=\\\\ =\frac{e^{\frac{\pi}{2}}-1}{2}$