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Integral of d{x}:
Integral of x^1
Integral of sin(x)/cos(x)
Integral of -2*x
Integral of e^(x+2)
Identical expressions
e^(-2x)*sin2x
e to the power of ( minus 2x) multiply by sinus of 2x
e(-2x)*sin2x
e-2x*sin2x
e^(-2x)sin2x
e(-2x)sin2x
e-2xsin2x
e^-2xsin2x
e^(-2x)*sin2xdx
Similar expressions
e^(2x)*sin2x

Integral of e^(-2x)*sin2x dx

The solution

You have entered [src]

  1                  
  /                  
 |                   
 |   -2*x            
 |  e    *sin(2*x) dx
 |                   
/                    
0

$$\int\limits_{0}^{1} e^{- 2 x} \sin{\left(2 x \right)}\, dx$$

Integral(sin(2*x)/E^(2*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x$ .
  
  Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{e^{- u} \sin{\left(u \right)}}{4}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{e^{- u} \sin{\left(u \right)}}{2}\, du = \frac{\int e^{- u} \sin{\left(u \right)}\, du}{2}$
    1. Let $u = - u$ .
      
      Then let $du = - du$ and substitute $du$ :
      
      $\int e^{u} \sin{\left(u \right)}\, du$
      
      Use integration by parts, noting that the integrand eventually repeats itself.
      
      For the integrand $e^{u} \sin{\left(u \right)}$ :
      
      Let $u{\left(u \right)} = \sin{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - \int e^{u} \cos{\left(u \right)}\, du$ .
      
      For the integrand $e^{u} \cos{\left(u \right)}$ :
      
      Let $u{\left(u \right)} = \cos{\left(u \right)}$ and let $\operatorname{dv}{\left(u \right)} = e^{u}$ .
      
      Then $\int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)} + \int \left(- e^{u} \sin{\left(u \right)}\right)\, du$ .
      
      Notice that the integrand has repeated itself, so move it to one side:
      
      $2 \int e^{u} \sin{\left(u \right)}\, du = e^{u} \sin{\left(u \right)} - e^{u} \cos{\left(u \right)}$
      
      Therefore,
      
      $\int e^{u} \sin{\left(u \right)}\, du = \frac{e^{u} \sin{\left(u \right)}}{2} - \frac{e^{u} \cos{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{e^{- u} \sin{\left(u \right)}}{2} - \frac{e^{- u} \cos{\left(u \right)}}{2}$
    So, the result is: $- \frac{e^{- u} \sin{\left(u \right)}}{4} - \frac{e^{- u} \cos{\left(u \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{e^{- 2 x} \sin{\left(2 x \right)}}{4} - \frac{e^{- 2 x} \cos{\left(2 x \right)}}{4}$
Method #2
1. Use integration by parts, noting that the integrand eventually repeats itself.
  1. For the integrand $e^{- 2 x} \sin{\left(2 x \right)}$ :
    
    Let $u{\left(x \right)} = \sin{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$ .
    
    Then $\int e^{- 2 x} \sin{\left(2 x \right)}\, dx = - \int \left(- e^{- 2 x} \cos{\left(2 x \right)}\right)\, dx - \frac{e^{- 2 x} \sin{\left(2 x \right)}}{2}$ .
  2. For the integrand $- e^{- 2 x} \cos{\left(2 x \right)}$ :
    
    Let $u{\left(x \right)} = - \cos{\left(2 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{- 2 x}$ .
    
    Then $\int e^{- 2 x} \sin{\left(2 x \right)}\, dx = \int \left(- e^{- 2 x} \sin{\left(2 x \right)}\right)\, dx - \frac{e^{- 2 x} \sin{\left(2 x \right)}}{2} - \frac{e^{- 2 x} \cos{\left(2 x \right)}}{2}$ .
  3. Notice that the integrand has repeated itself, so move it to one side:
    
    $2 \int e^{- 2 x} \sin{\left(2 x \right)}\, dx = - \frac{e^{- 2 x} \sin{\left(2 x \right)}}{2} - \frac{e^{- 2 x} \cos{\left(2 x \right)}}{2}$
    
    Therefore,
    
    $\int e^{- 2 x} \sin{\left(2 x \right)}\, dx = - \frac{e^{- 2 x} \sin{\left(2 x \right)}}{4} - \frac{e^{- 2 x} \cos{\left(2 x \right)}}{4}$
Now simplify:

$- \frac{\sqrt{2} e^{- 2 x} \sin{\left(2 x + \frac{\pi}{4} \right)}}{4}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                       
 |                                   -2*x    -2*x         
 |  -2*x                   cos(2*x)*e       e    *sin(2*x)
 | e    *sin(2*x) dx = C - -------------- - --------------
 |                               4                4       
/

$$\int e^{- 2 x} \sin{\left(2 x \right)}\, dx = C - \frac{e^{- 2 x} \sin{\left(2 x \right)}}{4} - \frac{e^{- 2 x} \cos{\left(2 x \right)}}{4}$$

Simplify

The answer [src]

            -2    -2       
1   cos(2)*e     e  *sin(2)
- - ---------- - ----------
4       4            4

$$- \frac{\sin{\left(2 \right)}}{4 e^{2}} - \frac{\cos{\left(2 \right)}}{4 e^{2}} + \frac{1}{4}$$

            -2    -2       
1   cos(2)*e     e  *sin(2)
- - ---------- - ----------
4       4            4

$$- \frac{\sin{\left(2 \right)}}{4 e^{2}} - \frac{\cos{\left(2 \right)}}{4 e^{2}} + \frac{1}{4}$$

Упростить

Numerical answer [src]

0.233314831296588

0.233314831296588

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

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How to use it?

Identical expressions

Similar expressions

Integral of e^(-2x)*sin2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2