Integral of e^(3*x)*sin(2*x) dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫2e3xsin(x)cos(x)dx=2∫e3xsin(x)cos(x)dx
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=sin(x)cos(x) and let dv(x)=e3x.
Then du(x)=−sin2(x)+cos2(x).
To find v(x):
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Let u=3x.
Then let du=3dx and substitute 3du:
∫3eudu
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The integral of a constant times a function is the constant times the integral of the function:
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The integral of the exponential function is itself.
∫eudu=eu
So, the result is: 3eu
Now substitute u back in:
3e3x
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫3(−sin2(x)+cos2(x))e3xdx=3∫(−sin2(x)+cos2(x))e3xdx
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Rewrite the integrand:
(−sin2(x)+cos2(x))e3x=−e3xsin2(x)+e3xcos2(x)
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Integrate term-by-term:
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The integral of a constant times a function is the constant times the integral of the function:
∫(−e3xsin2(x))dx=−∫e3xsin2(x)dx
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Don't know the steps in finding this integral.
But the integral is
3911e3xsin2(x)−132e3xsin(x)cos(x)+392e3xcos2(x)
So, the result is: −3911e3xsin2(x)+132e3xsin(x)cos(x)−392e3xcos2(x)
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Don't know the steps in finding this integral.
But the integral is
392e3xsin2(x)+132e3xsin(x)cos(x)+3911e3xcos2(x)
The result is: −133e3xsin2(x)+134e3xsin(x)cos(x)+133e3xcos2(x)
So, the result is: −13e3xsin2(x)+394e3xsin(x)cos(x)+13e3xcos2(x)
So, the result is: 132e3xsin2(x)+136e3xsin(x)cos(x)−132e3xcos2(x)
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Now simplify:
13(3sin(2x)−2cos(2x))e3x
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Add the constant of integration:
13(3sin(2x)−2cos(2x))e3x+constant
The answer is:
13(3sin(2x)−2cos(2x))e3x+constant
The answer (Indefinite)
[src]
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| 2 3*x 2 3*x 3*x
| 3*x 2*cos (x)*e 2*sin (x)*e 6*cos(x)*e *sin(x)
| E *sin(2*x) dx = C - -------------- + -------------- + --------------------
| 13 13 13
/
∫e3xsin(2x)dx=C+132e3xsin2(x)+136e3xsin(x)cos(x)−132e3xcos2(x)
The graph
⟨−∞,∞⟩
=
⟨−∞,∞⟩
3.06826157076743e+13000078344468995218
3.06826157076743e+13000078344468995218
Use the examples entering the upper and lower limits of integration.