Integral of exp(2x)*sin(3x) dx
The solution
Detail solution
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Use integration by parts, noting that the integrand eventually repeats itself.
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For the integrand e2xsin(3x):
Let u(x)=sin(3x) and let dv(x)=e2x.
Then ∫e2xsin(3x)dx=2e2xsin(3x)−∫23e2xcos(3x)dx.
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For the integrand 23e2xcos(3x):
Let u(x)=23cos(3x) and let dv(x)=e2x.
Then ∫e2xsin(3x)dx=2e2xsin(3x)−43e2xcos(3x)+∫(−49e2xsin(3x))dx.
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Notice that the integrand has repeated itself, so move it to one side:
413∫e2xsin(3x)dx=2e2xsin(3x)−43e2xcos(3x)
Therefore,
∫e2xsin(3x)dx=132e2xsin(3x)−133e2xcos(3x)
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Now simplify:
13(2sin(3x)−3cos(3x))e2x
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Add the constant of integration:
13(2sin(3x)−3cos(3x))e2x+constant
The answer is:
13(2sin(3x)−3cos(3x))e2x+constant
The answer (Indefinite)
[src]
/
| 2*x 2*x
| 2*x 3*cos(3*x)*e 2*e *sin(3*x)
| e *sin(3*x) dx = C - --------------- + ---------------
| 13 13
/
13e2x(2sin(3x)−3cos(3x))
2 2
3 3*cos(3)*e 2*e *sin(3)
-- - ----------- + -----------
13 13 13
132e2sin3−3e2cos3+133
=
2 2
3 3*cos(3)*e 2*e *sin(3)
-- - ----------- + -----------
13 13 13
132e2sin(3)+133−133e2cos(3)
Use the examples entering the upper and lower limits of integration.