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e^(-x)sinx
Solving integrals/ e^(-x)sinx

Integral of e^(-x)sinx dx

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The solution

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∫01e−xsin⁡(x) dx\int\limits_{0}^{1} e^{- x} \sin{\left(x \right)}\, dx 
Integral(E^(-x)*sin(x), (x, 0, 1))
Detail solution

  1. Let u=−xu = - x.

    Then let du=−dxdu = - dx and substitute dudu:

    ∫eusin⁡(u) du\int e^{u} \sin{\left(u \right)}\, du

    1. Use integration by parts, noting that the integrand eventually repeats itself.

      1. For the integrand eusin⁡(u)e^{u} \sin{\left(u \right)}:

        Let u(u)=sin⁡(u)u{\left(u \right)} = \sin{\left(u \right)} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

        Then ∫eusin⁡(u) du=eusin⁡(u)−∫eucos⁡(u) du\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - \int e^{u} \cos{\left(u \right)}\, du.

      2. For the integrand eucos⁡(u)e^{u} \cos{\left(u \right)}:

        Let u(u)=cos⁡(u)u{\left(u \right)} = \cos{\left(u \right)} and let dv⁡(u)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

        Then ∫eusin⁡(u) du=eusin⁡(u)−eucos⁡(u)+∫(−eusin⁡(u)) du\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)} + \int \left(- e^{u} \sin{\left(u \right)}\right)\, du.

      3. Notice that the integrand has repeated itself, so move it to one side:

        2∫eusin⁡(u) du=eusin⁡(u)−eucos⁡(u)2 \int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)}

        Therefore,

        ∫eusin⁡(u) du=eusin⁡(u)2−eucos⁡(u)2\int e^{u} \sin{\left(u \right)}\, du = \frac{e^{u} \sin{\left(u \right)}}{2} - \frac{e^{u} \cos{\left(u \right)}}{2}

    Now substitute uu back in:

    −e−xsin⁡(x)2−e−xcos⁡(x)2- \frac{e^{- x} \sin{\left(x \right)}}{2} - \frac{e^{- x} \cos{\left(x \right)}}{2}

  2. Now simplify:

    −2e−xsin⁡(x+π4)2- \frac{\sqrt{2} e^{- x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}

  3. Add the constant of integration:

    −2e−xsin⁡(x+π4)2+constant- \frac{\sqrt{2} e^{- x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}+ \mathrm{constant}

The answer is:

−2e−xsin⁡(x+π4)2+constant- \frac{\sqrt{2} e^{- x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                           
 |                             -x    -x       
 |  -x                 cos(x)*e     e  *sin(x)
 | E  *sin(x) dx = C - ---------- - ----------
 |                         2            2     
/
∫e−xsin⁡(x) dx=C−e−xsin⁡(x)2−e−xcos⁡(x)2\int e^{- x} \sin{\left(x \right)}\, dx = C - \frac{e^{- x} \sin{\left(x \right)}}{2} - \frac{e^{- x} \cos{\left(x \right)}}{2}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.901.0-1.0
Plot the graph f(x)
The answer [src] 
            -1    -1       
1   cos(1)*e     e  *sin(1)
- - ---------- - ----------
2       2            2
−sin⁡(1)2e−cos⁡(1)2e+12- \frac{\sin{\left(1 \right)}}{2 e} - \frac{\cos{\left(1 \right)}}{2 e} + \frac{1}{2} 
=
= 
            -1    -1       
1   cos(1)*e     e  *sin(1)
- - ---------- - ----------
2       2            2
−sin⁡(1)2e−cos⁡(1)2e+12- \frac{\sin{\left(1 \right)}}{2 e} - \frac{\cos{\left(1 \right)}}{2 e} + \frac{1}{2} 
1/2 - cos(1)*exp(-1)/2 - exp(-1)*sin(1)/2
Numerical answer [src] 
0.245837007000237
0.245837007000237
The graph
Integral of e^(-x)sinx dx

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

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