Question

Hello! How should i solve this
∫e^(-t)/2) *sin⁡(3t)dt
I know what is the result but, i dont know how to get there.
Thanks

Answer 1

, is that

\[\int e^{-t/2} sin(3t)dt\]

Answer 2

?

Answer 3

yes

Answer 4

by parts

Answer 5

i have tried, but it always repeat cos(3t)/3, sin(3t)/3 and -2*e^(-1/2) or?

Answer 6

a = b - a

a+a = b

2a = b

a = b/2

Answer 7

if you get a repeat; use it ...

Answer 8

write it up till your first repeat ... and lets see if we can make this more coherent ...

Answer 9

sin(3t)=u, e^(-t/2)*dt = du

-2*e^(-t/2)*sin(3t) + ∫2*e^(-t/2)*cos(3t)/3*dt

Answer 10

e^(-t/2)*dt = dv

Answer 11

good, and one more time should get you back around to the sin e stuff

Answer 12

yes

Answer 13

\[\int e^{-t/2} sin(3t)dt=e^{-t/2}*3cos(3t)-\int3cos(3t)*\frac{-e^{-t/2}}{2}dt\]
\[\int e^{-t/2} sin(3t)dt3=e^{-t/2}cos(3t)+\frac{3}{2}\left(\int e^{-t/2}cos(3t)dt\right)\]
\[\int e^{-t/2} sin(3t)dt=...+\frac{3}{2}\left(-3e^{-t/2}sin(3t)-\frac{3}{2}\int e^{-t/2}sin(3t)dt\right)\]
i believe thats going up correctly

Answer 14

\[\sin (3 t)=\text{Complex}\left[e^{3 i t}\right]\]
\[e^{-\frac{t}{2}} e^{3 i t}=e^{\left(-\frac{1}{2}+3 i\right) t}\]

\[\int e^{\left(-\frac{1}{2}+3 i\right) t} \, dt\]

\[\left(-\frac{2}{37}-\frac{12 i}{37}\right) e^{\left(-\frac{1}{2}+3 i\right) t}=\frac{12}{37} e^{-\frac{t}{2}} \sin (3 t)+\frac{1}{37} (-2) e^{-\frac{t}{2}} \cos (3 t)+\]
\[ i \left(\frac{1}{37} (-12) e^{-\frac{t}{2}} \cos (3 t)-\frac{2}{37} e^{-\frac{t}{2}} \sin (3 t)\right)\]

only taking the complex part

=\[ \left(\frac{1}{37} (-12) e^{-\frac{t}{2}} \cos (3 t)-\frac{2}{37} e^{-\frac{t}{2}} \sin (3 t)\right)\]

Answer 15

\[\int e^{-t/2} sin(3t)dt=...-\frac{9}{4} \int e^{-t/2}sin(3t)dt\]
\[\int e^{-t/2} sin(3t)dt + \frac{9}{4} \int e^{-t/2}sin(3t)dt=...\]
\[(1+ \frac{9}{4})\int e^{-t/2} sin(3t)dt=...\]
\[\int e^{-t/2} sin(3t)dt=\frac{...}{1+\frac{9}{4}}\]

Answer 16

but will this ever end?

Answer 17

that is the end.

Answer 18

get to the point that it repeats itself; and then use that repetition to solve it

Answer 19

ok, i will try

Answer 20

thanks

Answer 21

spose:\[A=\int e^{-t/2}sin(3t)\ dt\]
and you get to the point\[A = f(t)-kA\]
this is the repetition, so use it. add kA to each side
\[A+kA = f(t)-kA+kA\]
\[A+kA = f(t)\cancel{-kA+kA}\]
\[(1+k)A = f(t)\]
\[A = \frac{f(t)}{1+k}\]