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Integral of e^(3*x)*sin(2*x) dx

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The solution

You have entered [src]
 oo                 
  /                 
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 |   3*x            
 |  E   *sin(2*x) dx
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-oo                 
e3xsin(2x)dx\int\limits_{-\infty}^{\infty} e^{3 x} \sin{\left(2 x \right)}\, dx
Integral(E^(3*x)*sin(2*x), (x, -oo, oo))
Detail solution
  1. The integral of a constant times a function is the constant times the integral of the function:

    2e3xsin(x)cos(x)dx=2e3xsin(x)cos(x)dx\int 2 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}\, dx

    1. Use integration by parts:

      udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=sin(x)cos(x)u{\left(x \right)} = \sin{\left(x \right)} \cos{\left(x \right)} and let dv(x)=e3x\operatorname{dv}{\left(x \right)} = e^{3 x}.

      Then du(x)=sin2(x)+cos2(x)\operatorname{du}{\left(x \right)} = - \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}.

      To find v(x)v{\left(x \right)}:

      1. Let u=3xu = 3 x.

        Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

        eu3du\int \frac{e^{u}}{3}\, du

        1. The integral of a constant times a function is the constant times the integral of the function:

          False\text{False}

          1. The integral of the exponential function is itself.

            eudu=eu\int e^{u}\, du = e^{u}

          So, the result is: eu3\frac{e^{u}}{3}

        Now substitute uu back in:

        e3x3\frac{e^{3 x}}{3}

      Now evaluate the sub-integral.

    2. The integral of a constant times a function is the constant times the integral of the function:

      (sin2(x)+cos2(x))e3x3dx=(sin2(x)+cos2(x))e3xdx3\int \frac{\left(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) e^{3 x}}{3}\, dx = \frac{\int \left(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) e^{3 x}\, dx}{3}

      1. Rewrite the integrand:

        (sin2(x)+cos2(x))e3x=e3xsin2(x)+e3xcos2(x)\left(- \sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right) e^{3 x} = - e^{3 x} \sin^{2}{\left(x \right)} + e^{3 x} \cos^{2}{\left(x \right)}

      2. Integrate term-by-term:

        1. The integral of a constant times a function is the constant times the integral of the function:

          (e3xsin2(x))dx=e3xsin2(x)dx\int \left(- e^{3 x} \sin^{2}{\left(x \right)}\right)\, dx = - \int e^{3 x} \sin^{2}{\left(x \right)}\, dx

          1. Don't know the steps in finding this integral.

            But the integral is

            11e3xsin2(x)392e3xsin(x)cos(x)13+2e3xcos2(x)39\frac{11 e^{3 x} \sin^{2}{\left(x \right)}}{39} - \frac{2 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{13} + \frac{2 e^{3 x} \cos^{2}{\left(x \right)}}{39}

          So, the result is: 11e3xsin2(x)39+2e3xsin(x)cos(x)132e3xcos2(x)39- \frac{11 e^{3 x} \sin^{2}{\left(x \right)}}{39} + \frac{2 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{13} - \frac{2 e^{3 x} \cos^{2}{\left(x \right)}}{39}

        1. Don't know the steps in finding this integral.

          But the integral is

          2e3xsin2(x)39+2e3xsin(x)cos(x)13+11e3xcos2(x)39\frac{2 e^{3 x} \sin^{2}{\left(x \right)}}{39} + \frac{2 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{13} + \frac{11 e^{3 x} \cos^{2}{\left(x \right)}}{39}

        The result is: 3e3xsin2(x)13+4e3xsin(x)cos(x)13+3e3xcos2(x)13- \frac{3 e^{3 x} \sin^{2}{\left(x \right)}}{13} + \frac{4 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{13} + \frac{3 e^{3 x} \cos^{2}{\left(x \right)}}{13}

      So, the result is: e3xsin2(x)13+4e3xsin(x)cos(x)39+e3xcos2(x)13- \frac{e^{3 x} \sin^{2}{\left(x \right)}}{13} + \frac{4 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{39} + \frac{e^{3 x} \cos^{2}{\left(x \right)}}{13}

    So, the result is: 2e3xsin2(x)13+6e3xsin(x)cos(x)132e3xcos2(x)13\frac{2 e^{3 x} \sin^{2}{\left(x \right)}}{13} + \frac{6 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{13} - \frac{2 e^{3 x} \cos^{2}{\left(x \right)}}{13}

  2. Now simplify:

    (3sin(2x)2cos(2x))e3x13\frac{\left(3 \sin{\left(2 x \right)} - 2 \cos{\left(2 x \right)}\right) e^{3 x}}{13}

  3. Add the constant of integration:

    (3sin(2x)2cos(2x))e3x13+constant\frac{\left(3 \sin{\left(2 x \right)} - 2 \cos{\left(2 x \right)}\right) e^{3 x}}{13}+ \mathrm{constant}


The answer is:

(3sin(2x)2cos(2x))e3x13+constant\frac{\left(3 \sin{\left(2 x \right)} - 2 \cos{\left(2 x \right)}\right) e^{3 x}}{13}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                                             
 |                             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+2e3xsin2(x)13+6e3xsin(x)cos(x)132e3xcos2(x)13\int e^{3 x} \sin{\left(2 x \right)}\, dx = C + \frac{2 e^{3 x} \sin^{2}{\left(x \right)}}{13} + \frac{6 e^{3 x} \sin{\left(x \right)} \cos{\left(x \right)}}{13} - \frac{2 e^{3 x} \cos^{2}{\left(x \right)}}{13}
The graph
-0.010-0.008-0.006-0.004-0.0020.0100.0000.0020.0040.0060.0080.02-0.02
The answer [src]
<-oo, oo>
,\left\langle -\infty, \infty\right\rangle
=
=
<-oo, oo>
,\left\langle -\infty, \infty\right\rangle
AccumBounds(-oo, oo)
Numerical answer [src]
3.06826157076743e+13000078344468995218
3.06826157076743e+13000078344468995218

    Use the examples entering the upper and lower limits of integration.