Integral of e^(5x)×sin3x 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 e5xsin(3x):
Let u(x)=sin(3x) and let dv(x)=e5x.
Then ∫e5xsin(3x)dx=5e5xsin(3x)−∫53e5xcos(3x)dx.
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For the integrand 53e5xcos(3x):
Let u(x)=53cos(3x) and let dv(x)=e5x.
Then ∫e5xsin(3x)dx=5e5xsin(3x)−253e5xcos(3x)+∫(−259e5xsin(3x))dx.
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Notice that the integrand has repeated itself, so move it to one side:
2534∫e5xsin(3x)dx=5e5xsin(3x)−253e5xcos(3x)
Therefore,
∫e5xsin(3x)dx=345e5xsin(3x)−343e5xcos(3x)
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Now simplify:
34(5sin(3x)−3cos(3x))e5x
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Add the constant of integration:
34(5sin(3x)−3cos(3x))e5x+constant
The answer is:
34(5sin(3x)−3cos(3x))e5x+constant
The answer (Indefinite)
[src]
/
| 5*x 5*x
| 5*x 3*cos(3*x)*e 5*e *sin(3*x)
| e *sin(3*x) dx = C - --------------- + ---------------
| 34 34
/
34e5x(5sin(3x)−3cos(3x))
The graph
5 5
3 3*cos(3)*e 5*e *sin(3)
-- - ----------- + -----------
34 34 34
345e5sin3−3e5cos3+343
=
5 5
3 3*cos(3)*e 5*e *sin(3)
-- - ----------- + -----------
34 34 34
343+345e5sin(3)−343e5cos(3)
Use the examples entering the upper and lower limits of integration.