∫ ( e 2 θ ) s i n 3 θ d θ {\displaystyle \int \left(e^{2{\theta }}\right)sin{3\theta }d{\theta }}
First:
u = sin ( 3 θ ) {\displaystyle u={\sin {(3\theta )}}} d u = 3 cos ( 3 θ ) d θ {\displaystyle du={3\cos {(3\theta )}}d{\theta }}
d v = ( e 2 θ ) {\displaystyle dv=\left(e^{2{\theta }}\right)} v = 1 2 ( e 2 θ ) {\displaystyle v={\frac {1}{2}}\left(e^{2{\theta }}\right)}
Then: I = ∫ ( e 2 θ ) sin 3 θ d θ = 1 2 ( e 2 θ ) ∗ sin ( 3 θ ) − 3 2 ∫ ( e 2 θ ) cos 3 θ d θ {\displaystyle 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 θ ) {\displaystyle U={\cos {(3\theta )}}}
d U = 3 sin ( 3 θ ) d θ {\displaystyle dU={3\sin {(3\theta )}}d{\theta }}
d V = ( e 2 θ ) d θ {\displaystyle dV=\left(e^{2{\theta }}\right)d{\theta }}
V = 1 2 ( e 2 θ ) {\displaystyle V={\frac {1}{2}}\left(e^{2{\theta }}\right)} to get
∫ ( e 2 θ ) cos 3 θ d θ = 1 2 ∫ ( e 2 θ ) cos 3 θ d θ + 3 2 ∫ ( e 2 θ ) sin 3 θ d θ {\displaystyle \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 = 1 2 ( e 2 θ ) sin 3 θ − 3 4 ( e 2 θ ) cos 3 θ − 9 4 ∫ ( e 2 θ ) s i n 3 θ d θ {\displaystyle 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 }}
= 1 2 ( e 2 θ ) s i n 3 θ − 3 4 ( e 2 θ ) c o s 3 θ − 9 4 I {\displaystyle ={\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: 13 4 I = 1 2 ( e 2 θ ) s i n 3 θ − 3 4 ( e 2 θ ) c o s 3 θ + C {\displaystyle {\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 = 1 13 ( e 2 θ ) 2 s i n 3 θ − 3 c o s 3 θ + C {\displaystyle I={\frac {1}{13}}\left(e^{2{\theta }}\right)2sin{3\theta }-{3cos3\theta }+C} ,
Where C = 4 13 C {\displaystyle C={\frac {4}{13}}C}
to get
substituting in the previous formula gives
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