\(\displaystyle{ \int x ^{2}cos3xdx= \begin{vmatrix} u=x ^{2},u'=2x \\ v'=cos3x,v= \frac{1}{3}sin3x \end{vmatrix} = \frac{1}{3} x ^{2}sin3x - \frac{1}{3}\int 2xsin3xdx = \begin{vmatrix} u=2x, u'=2 \\v'=sin3x,v=- \frac{1}{3}cos3x \end{vmatrix} = \frac{1}{3}x ^{2}sin3x + \frac{2}{9}xcos3x + \frac{2}{9} t cos3x dx = \frac{1}{3}x ^{2} sinx3x + \frac{2}{9} xcos3x- \frac{2}{27}sin3x +C}\)
\(\displaystyle{ \int xe ^{3x}dx = \begin{vmatrix} u=x,u'=1 \\v'=e ^{3x},v= \frac{1}{3}e ^{3x} \end{vmatrix} = \frac{1}{3}xe ^{3x}- \frac{1}{3}\int e ^{3x}dx= \frac{1}{3}xe ^{3x}- \frac{1}{9}e ^{3x} +C}\)
\(\displaystyle{ \int \frac{1}{1-cos} = t dx - t \frac{1}{cosx}dx= x - lnsinx +C}\)
poprosze o sprawdzenie. dzieki.