7.1 Integration By Parts/17

$\int \left(e^{2{\theta }}\right)sin{3\theta }d{\theta }$

First:

$u={\sin {(3\theta )}}$ $du={3\cos {(3\theta )}}d{\theta }$

$dv=\left(e^{2{\theta }}\right)$ $v={\frac {1}{2}}\left(e^{2{\theta }}\right)$

Then: $I=\int \left(e^{2{\theta }}\right)\sin {3\theta }d{\theta }={\frac {1}{2}}\left(e^{2{\theta }}\right)*{\sin {(3\theta )}}-{\frac {3}{2}}\int \left(e^{2{\theta }}\right)\cos {3\theta }d{\theta }$

Next: let

$U={\cos {(3\theta )}}$

$dU={3\sin {(3\theta )}}d{\theta }$

$dV=\left(e^{2{\theta }}\right)d{\theta }$

$V={\frac {1}{2}}\left(e^{2{\theta }}\right)$ to get

$\int \left(e^{2{\theta }}\right)\cos {3\theta }d{\theta }={\frac {1}{2}}\int \left(e^{2{\theta }}\right)\cos {3\theta }d{\theta }+{\frac {3}{2}}\int \left(e^{2{\theta }}\right)\sin {3\theta }d{\theta }$ substituting in the previous formula gives

$I={\frac {1}{2}}\left(e^{2{\theta }}\right)\sin {3\theta }-{\frac {3}{4}}\left(e^{2{\theta }}\right)\cos {3\theta }-{\frac {9}{4}}\int \left(e^{2{\theta }}\right)sin{3\theta }d{\theta }$

$={\frac {1}{2}}\left(e^{2{\theta }}\right)sin{3\theta }-{\frac {3}{4}}\left(e^{2{\theta }}\right)cos{3\theta }-{\frac {9}{4}}I$

Then: ${\frac {13}{4}}I={\frac {1}{2}}\left(e^{2{\theta }}\right)sin{3\theta }-{\frac {3}{4}}\left(e^{2{\theta }}\right)cos{3\theta }+C$,

Hence,$I={\frac {1}{13}}\left(e^{2{\theta }}\right)2sin{3\theta }-{3cos3\theta }+C$,

Where $C={\frac {4}{13}}C$