Вычислить интеграл e^3x cos^2 * x dx Распишите поподробнее, если возможно.

Вот как вижу так и решаю.
$\int e^{3x}cos^2xdx=\frac{1}{3}e^{3x}cos^2x+\frac{1}{3}\int e^{3x}sin2xdx=\\=\frac{1}{3}e^{3x}cos^2x+\frac{1}{9}e^{3x}sin2x-\frac{4}{9}\int e^{3x}cos^2xdx+\frac{2}{27}e^{3x}\\\\\frac{13}{9}\int e^{3x}cos^2xdx=\frac{1}{3}e^{3x}cos^2x+\frac{1}{9}e^{3x}sin2x+\frac{2}{27}e^{3x}\\\\\int e^{3x}cos^2xdx=\frac{3}{13}e^{3x}cos^2x+\frac{1}{13}e^{3x}sin2x+\frac{2}{39}e^{3x}+C\\\\u=cos^2x=\ \textgreater \ du=-sin2xdx\\dv=e^{3x}dx=\ \textgreater \ v=\frac{1}{3}e^{3x}$
$\int e^{3x}sin2xdx=\frac{1}{3}e^{3x}sin2x-\frac{4}{3}\int e^{3x}cos^2xdx+\frac{2}{3}\int e^{3x}dx=\\=\frac{1}{3}e^{3x}sin2x-\frac{4}{3}\int e^{3x}cos^2xdx+\frac{2}{9}e^{3x}\\u=sin2x=\ \textgreater \ du=(4cos^2x-2)dx\\dv=e^{3x}dx=\ \textgreater \ v=\frac{1}{3}e^{3x}$

$(\frac{3}{13}e^{3x}cos^2x+\frac{1}{13}e^{3x}sin2x+\frac{2}{39}e^{3x}+C)'=\\=\frac{3}{13}(3e^{3x}cos^2x-e^{3x}sin2x)+\frac{1}{13}(3e^{3x}sin2x+4e^{3x}cos^2x-2e^{3x})+\\+\frac{2}{13}e^{3x}=\frac{9}{13}e^{3x}cos^2x-\frac{3}{13}e^{3x}sin2x+\frac{3}{13}e^{3x}sin2x+\frac{4}{13}e^{3x}cos^2x-\\-\frac{2}{13}e^{3x}+\frac{2}{13}e^{3x}=e^{3x}cos^2x$