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Solving integrals/ e^(-2x)*sin(5x)

Integral of e^(-2x)*sin(5x) dx

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The solution

You have entered [src] 
 oo                  
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 |   -2*x            
 |  e    *sin(5*x) dx
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∫0∞e−2xsin⁡(5x) dx\int\limits_{0}^{\infty} e^{- 2 x} \sin{\left(5 x \right)}\, dx
Simplify
Detail solution

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

    1. For the integrand e−2xsin⁡(5x)e^{- 2 x} \sin{\left(5 x \right)}:

      Let u(x)=sin⁡(5x)u{\left(x \right)} = \sin{\left(5 x \right)} and let dv⁡(x)=e−2x\operatorname{dv}{\left(x \right)} = e^{- 2 x}.

      Then ∫e−2xsin⁡(5x) dx=−∫(−5e−2xcos⁡(5x)2) dx−e−2xsin⁡(5x)2\int e^{- 2 x} \sin{\left(5 x \right)}\, dx = - \int \left(- \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{2}\right)\, dx - \frac{e^{- 2 x} \sin{\left(5 x \right)}}{2}.

    2. For the integrand −5e−2xcos⁡(5x)2- \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{2}:

      Let u(x)=−5cos⁡(5x)2u{\left(x \right)} = - \frac{5 \cos{\left(5 x \right)}}{2} and let dv⁡(x)=e−2x\operatorname{dv}{\left(x \right)} = e^{- 2 x}.

      Then ∫e−2xsin⁡(5x) dx=∫(−25e−2xsin⁡(5x)4) dx−e−2xsin⁡(5x)2−5e−2xcos⁡(5x)4\int e^{- 2 x} \sin{\left(5 x \right)}\, dx = \int \left(- \frac{25 e^{- 2 x} \sin{\left(5 x \right)}}{4}\right)\, dx - \frac{e^{- 2 x} \sin{\left(5 x \right)}}{2} - \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{4}.

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

      29∫e−2xsin⁡(5x) dx4=−e−2xsin⁡(5x)2−5e−2xcos⁡(5x)4\frac{29 \int e^{- 2 x} \sin{\left(5 x \right)}\, dx}{4} = - \frac{e^{- 2 x} \sin{\left(5 x \right)}}{2} - \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{4}

      Therefore,

      ∫e−2xsin⁡(5x) dx=−2e−2xsin⁡(5x)29−5e−2xcos⁡(5x)29\int e^{- 2 x} \sin{\left(5 x \right)}\, dx = - \frac{2 e^{- 2 x} \sin{\left(5 x \right)}}{29} - \frac{5 e^{- 2 x} \cos{\left(5 x \right)}}{29}

  2. Now simplify:

    −(2sin⁡(5x)+5cos⁡(5x))e−2x29- \frac{\left(2 \sin{\left(5 x \right)} + 5 \cos{\left(5 x \right)}\right) e^{- 2 x}}{29}

  3. Add the constant of integration:

    −(2sin⁡(5x)+5cos⁡(5x))e−2x29+constant- \frac{\left(2 \sin{\left(5 x \right)} + 5 \cos{\left(5 x \right)}\right) e^{- 2 x}}{29}+ \mathrm{constant}

The answer is:

−(2sin⁡(5x)+5cos⁡(5x))e−2x29+constant- \frac{\left(2 \sin{\left(5 x \right)} + 5 \cos{\left(5 x \right)}\right) e^{- 2 x}}{29}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                           
 |                                     -2*x      -2*x         
 |  -2*x                   5*cos(5*x)*e       2*e    *sin(5*x)
 | e    *sin(5*x) dx = C - ---------------- - ----------------
 |                                29                 29       
/
e−2 x (−2 sin⁡(5 x)−5 cos⁡(5 x))29{{e^ {- 2\,x }\,\left(-2\,\sin \left(5\,x\right)-5\,\cos \left(5\,x \right)\right)}\over{29}}
Simplify
The answer [src] 
5/29
529\frac{5}{29} 
=
= 
5/29
529\frac{5}{29}
Simplify

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

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