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Identical expressions
e^(-x)cos(nx)
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e(-x)cos(nx)
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Similar expressions
e^(x)cos(nx)

Integral of e^(-x)cos(nx) dx

The solution

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  1                
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 |   -x            
 |  e  *cos(n*x) dx
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0

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

Simplify

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \cos{\left(n x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{- x}$ .

Then $\operatorname{du}{\left(x \right)} = - n \sin{\left(n x \right)}$ .

To find $v{\left(x \right)}$ :
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = - x$ .
    
    Then let $du = - dx$ and substitute $- du$ :
    
    $\int e^{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- e^{u}\right)\, du = - \int e^{u}\, du$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $- e^{u}$
    Now substitute $u$ back in:
    
    $- e^{- x}$
  Method #2
  1. Let $u = e^{- x}$ .
    
    Then let $du = - e^{- x} dx$ and substitute $- du$ :
    
    $\int 1\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(-1\right)\, du = - \int 1\, du$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $- u$
    Now substitute $u$ back in:
    
    $- e^{- x}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int n e^{- x} \sin{\left(n x \right)}\, dx = n \int e^{- x} \sin{\left(n x \right)}\, dx$
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \sin{\left(n x \right)}$ and let $\operatorname{dv}{\left(x \right)} = e^{- x}$ .
  
  Then $\operatorname{du}{\left(x \right)} = n \cos{\left(n x \right)}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = - x$ .
    
    Then let $du = - dx$ and substitute $- du$ :
    
    $\int e^{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- e^{u}\right)\, du = - \int e^{u}\, du$
      1. The integral of the exponential function is itself.
        
        $\int e^{u}\, du = e^{u}$
      So, the result is: $- e^{u}$
    Now substitute $u$ back in:
    
    $- e^{- x}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- n e^{- x} \cos{\left(n x \right)}\right)\, dx = - n \int e^{- x} \cos{\left(n x \right)}\, dx$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $\begin{cases} \frac{x e^{- x} \sinh{\left(x \right)}}{2} + \frac{x e^{- x} \cosh{\left(x \right)}}{2} - \frac{e^{- x} \cosh{\left(x \right)}}{2} & \text{for}\: n = - i \vee n = i \\\frac{n \sin{\left(n x \right)}}{n^{2} e^{x} + e^{x}} - \frac{\cos{\left(n x \right)}}{n^{2} e^{x} + e^{x}} & \text{otherwise} \end{cases}$
  So, the result is: $- n \left(\begin{cases} \frac{x e^{- x} \sinh{\left(x \right)}}{2} + \frac{x e^{- x} \cosh{\left(x \right)}}{2} - \frac{e^{- x} \cosh{\left(x \right)}}{2} & \text{for}\: n = - i \vee n = i \\\frac{n \sin{\left(n x \right)}}{n^{2} e^{x} + e^{x}} - \frac{\cos{\left(n x \right)}}{n^{2} e^{x} + e^{x}} & \text{otherwise} \end{cases}\right)$
So, the result is: $n \left(n \left(\begin{cases} \frac{x e^{- x} \sinh{\left(x \right)}}{2} + \frac{x e^{- x} \cosh{\left(x \right)}}{2} - \frac{e^{- x} \cosh{\left(x \right)}}{2} & \text{for}\: n = - i \vee n = i \\\frac{n \sin{\left(n x \right)}}{n^{2} e^{x} + e^{x}} - \frac{\cos{\left(n x \right)}}{n^{2} e^{x} + e^{x}} & \text{otherwise} \end{cases}\right) - e^{- x} \sin{\left(n x \right)}\right)$
Now simplify:

$\begin{cases} - \left(\frac{n^{2} \left(x \sinh{\left(x \right)} + x \cosh{\left(x \right)} - \cosh{\left(x \right)}\right)}{2} - n \sin{\left(n x \right)} + \cos{\left(n x \right)}\right) e^{- x} & \text{for}\: n = - i \vee n = i \\\frac{\left(n \sin{\left(n x \right)} - \cos{\left(n x \right)}\right) e^{- x}}{n^{2} + 1} & \text{otherwise} \end{cases}$
Add the constant of integration:

$\begin{cases} - \left(\frac{n^{2} \left(x \sinh{\left(x \right)} + x \cosh{\left(x \right)} - \cosh{\left(x \right)}\right)}{2} - n \sin{\left(n x \right)} + \cos{\left(n x \right)}\right) e^{- x} & \text{for}\: n = - i \vee n = i \\\frac{\left(n \sin{\left(n x \right)} - \cos{\left(n x \right)}\right) e^{- x}}{n^{2} + 1} & \text{otherwise} \end{cases}+ \mathrm{constant}$

The answer is:

$\begin{cases} - \left(\frac{n^{2} \left(x \sinh{\left(x \right)} + x \cosh{\left(x \right)} - \cosh{\left(x \right)}\right)}{2} - n \sin{\left(n x \right)} + \cos{\left(n x \right)}\right) e^{- x} & \text{for}\: n = - i \vee n = i \\\frac{\left(n \sin{\left(n x \right)} - \cos{\left(n x \right)}\right) e^{- x}}{n^{2} + 1} & \text{otherwise} \end{cases}+ \mathrm{constant}$

The answer (Indefinite) [src]

                           /  //           -x              -x      -x                               \               \               
  /                        |  ||  cosh(x)*e     x*cosh(x)*e     x*e  *sinh(x)                       |               |               
 |                         |  ||- ----------- + ------------- + -------------  for Or(n = -I, n = I)|               |               
 |  -x                     |  ||       2              2               2                             |    -x         |             -x
 | e  *cos(n*x) dx = C - n*|n*|<                                                                    | - e  *sin(n*x)| - cos(n*x)*e  
 |                         |  ||             cos(n*x)    n*sin(n*x)                                 |               |               
/                          |  ||          - ---------- + ----------                  otherwise      |               |               
                           |  ||             2  x    x    2  x    x                                 |               |               
                           \  \\            n *e  + e    n *e  + e                                  /               /

$${{e^ {- x }\,\left(n\,\sin \left(n\,x\right)-\cos \left(n\,x\right) \right)}\over{n^2+1}}$$

Simplify

The answer [src]

  1       cos(n)    n*sin(n)
------ - -------- + --------
     2          2          2
1 + n    e + e*n    e + e*n

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

  1       cos(n)    n*sin(n)
------ - -------- + --------
     2          2          2
1 + n    e + e*n    e + e*n

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

Simplify

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

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Identical expressions

Similar expressions

Integral of e^(-x)cos(nx) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2