Dwie całki

[Lbubsazob]

1) \(\int e^{-3x} \sin 4x \mbox{d}x\)
2) \(\int 3x \cos \left( 5x-1\right)\mbox{d}x\)

[irena]

1)

\(P=\int\ e^{-3x}sin4x\ dx=*\\ \left(u'=e^{-3x},\ \ \ u=-\frac{1}{3}e^{-3x}\\v=sin4x,\ \ \ v'=4cos4x \right) \\*=-\frac{1}{3}e^{-3x}sin4x-\int(-\frac{1}{3}e^{-3x}\cdot4cos4x\ dx=-\frac{1}{3}e^{-3x}sin4x+\frac{4}{3}\int\ e^{-3x}cos4x\ dx=**\\ \left(u'=e^{-3x},\ \ \ u=-\frac{1}{3}e^{-3x}\\v=cos4x,\ \ \ v'=-4sin4x \right) \\**=-\frac{1}{3}e^{-3x}sin4x+\frac{4}{3}(-\frac{1}{3}e^{-3x}cos4x-\frac{4}{3}\int\ e^{-3x}sin4x\ dx)\)

\(P=-\frac{1}{3}e^{-3x}sin4x-\frac{4}{9}e^{-3x}cos4x-\frac{16}{9}P\\\frac{25}{9}P=-\frac{1}{3}e^{-3x}sin4x-\frac{4}{9}e^{-3x}cos4x\\P=-\frac{3}{25}e^{-3x}sin4x-\frac{4}{25}e^{-3x}cos4x\)

\(\int\ e^{-3x}sin4x\ dx=-\frac{e^{-3x}}{25}(3sin4x-4cos4x)+C\)

[irena]

2)
\(\int\ 3x\ cos(5x-1)dx=*\\ \left(u'=cos(5x-1),\ \ \ u=\frac{1}{5}sin(5x-1)\\v=3x,\ \ \ v'=3 \right) \\*=\frac{3}{5}x\ sin(5x-1)-\int\frac{3}{5}sin(5x-1)dx=\frac{3}{5}x\ sin(5x-1)-\frac{3}{5}\cdot(-\frac{1}{5})cos(5x-1)+C=\\=\frac{3}{5}x\ sin(5x-1)+\frac{3}{25}cos(5x-1)+C\)