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cos(z+ one)/z(z+ one)
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Similar expressions
cos(z+1)/z(z-1)
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Integral of cos(z+1)/z(z+1) dx

The solution

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  2                      
  /                      
 |                       
 |  cos(z + 1)           
 |  ----------*(z + 1) dz
 |      z                
 |                       
/                        
0

$$\int\limits_{0}^{2} \frac{\cos{\left(z + 1 \right)}}{z} \left(z + 1\right)\, dz$$

Integral((cos(z + 1)/z)*(z + 1), (z, 0, 2))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \frac{1}{z}$ .
  
  Then let $du = - \frac{dz}{z^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}\, du = - \int \frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}\, du$
    1. Rewrite the integrand:
      
      $\frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}} = \frac{\cos{\left(1 + \frac{1}{u} \right)}}{u} + \frac{\cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}$
    2. Integrate term-by-term:
      
      Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\cos{\left(u + 1 \right)}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u + 1 \right)}}{u}\, du = - \int \frac{\cos{\left(u + 1 \right)}}{u}\, du$
      
      CiRule(a=1, b=1, context=cos(_u + 1)/_u, symbol=_u)
      
      So, the result is: $- \cos{\left(1 \right)} \operatorname{Ci}{\left(u \right)} + \sin{\left(1 \right)} \operatorname{Si}{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \cos{\left(1 \right)} \operatorname{Ci}{\left(\frac{1}{u} \right)} + \sin{\left(1 \right)} \operatorname{Si}{\left(\frac{1}{u} \right)}$
      
      Let $u = 1 + \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \cos{\left(u \right)}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = - \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \sin{\left(1 + \frac{1}{u} \right)}$
      
      The result is: $- \sin{\left(1 + \frac{1}{u} \right)} - \cos{\left(1 \right)} \operatorname{Ci}{\left(\frac{1}{u} \right)} + \sin{\left(1 \right)} \operatorname{Si}{\left(\frac{1}{u} \right)}$
    So, the result is: $\sin{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(\frac{1}{u} \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(\frac{1}{u} \right)}$
  Now substitute $u$ back in:
  
  $\sin{\left(z + 1 \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(z \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(z \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\cos{\left(z + 1 \right)}}{z} \left(z + 1\right) = \frac{z \cos{\left(z + 1 \right)} + \cos{\left(z + 1 \right)}}{z}$
2. Let $u = \frac{1}{z}$ .
  
  Then let $du = - \frac{dz}{z^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}\, du = - \int \frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}\, du$
    1. Rewrite the integrand:
      
      $\frac{u \cos{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 + \frac{1}{u} \right)}}{u^{2}} = \frac{\cos{\left(1 + \frac{1}{u} \right)}}{u} + \frac{\cos{\left(1 + \frac{1}{u} \right)}}{u^{2}}$
    2. Integrate term-by-term:
      
      Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\cos{\left(u + 1 \right)}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u + 1 \right)}}{u}\, du = - \int \frac{\cos{\left(u + 1 \right)}}{u}\, du$
      
      CiRule(a=1, b=1, context=cos(_u + 1)/_u, symbol=_u)
      
      So, the result is: $- \cos{\left(1 \right)} \operatorname{Ci}{\left(u \right)} + \sin{\left(1 \right)} \operatorname{Si}{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \cos{\left(1 \right)} \operatorname{Ci}{\left(\frac{1}{u} \right)} + \sin{\left(1 \right)} \operatorname{Si}{\left(\frac{1}{u} \right)}$
      
      Let $u = 1 + \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \cos{\left(u \right)}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = - \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \sin{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \sin{\left(1 + \frac{1}{u} \right)}$
      
      The result is: $- \sin{\left(1 + \frac{1}{u} \right)} - \cos{\left(1 \right)} \operatorname{Ci}{\left(\frac{1}{u} \right)} + \sin{\left(1 \right)} \operatorname{Si}{\left(\frac{1}{u} \right)}$
    So, the result is: $\sin{\left(1 + \frac{1}{u} \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(\frac{1}{u} \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(\frac{1}{u} \right)}$
  Now substitute $u$ back in:
  
  $\sin{\left(z + 1 \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(z \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(z \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\cos{\left(z + 1 \right)}}{z} \left(z + 1\right) = \cos{\left(z + 1 \right)} + \frac{\cos{\left(z + 1 \right)}}{z}$
2. Integrate term-by-term:
  1. Let $u = z + 1$ .
    
    Then let $du = dz$ and substitute $du$ :
    
    $\int \cos{\left(u \right)}\, du$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    Now substitute $u$ back in:
    
    $\sin{\left(z + 1 \right)}$
  The result is: $\sin{\left(z + 1 \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(z \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(z \right)}$
Add the constant of integration:

$\sin{\left(z + 1 \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(z \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(z \right)}+ \mathrm{constant}$

The answer is:

$\sin{\left(z + 1 \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(z \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(z \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                    
 |                                                                     
 | cos(z + 1)                                                          
 | ----------*(z + 1) dz = C + Ci(z)*cos(1) - Si(z)*sin(1) + sin(1 + z)
 |     z                                                               
 |                                                                     
/

$$\int \frac{\cos{\left(z + 1 \right)}}{z} \left(z + 1\right)\, dz = C + \sin{\left(z + 1 \right)} + \cos{\left(1 \right)} \operatorname{Ci}{\left(z \right)} - \sin{\left(1 \right)} \operatorname{Si}{\left(z \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo - Si(2)*sin(1)

$$- \sin{\left(1 \right)} \operatorname{Si}{\left(2 \right)} + \infty$$

oo - Si(2)*sin(1)

$$- \sin{\left(1 \right)} \operatorname{Si}{\left(2 \right)} + \infty$$

oo - Si(2)*sin(1)

Numerical answer [src]

21.3130678400228

21.3130678400228

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

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Identical expressions

Similar expressions

Integral of cos(z+1)/z(z+1) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3