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Similar expressions
cosz/z(z^2-8)

Integral of cosz/z(z^2+8) dx

The solution

You have entered [src]

  0                   
  /                   
 |                    
 |  cos(z) / 2    \   
 |  ------*\z  + 8/ dz
 |    z               
 |                    
/                     
c

$$\int\limits_{c}^{0} \frac{\cos{\left(z \right)}}{z} \left(z^{2} + 8\right)\, dz$$

Integral((cos(z)/z)*(z^2 + 8), (z, c, 0))

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 $- du$ :
  
  $\int \left(- \frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}}\, du = - \int \frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}}\, du$
    1. Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{u^{2} \cos{\left(u \right)} + 8 \cos{\left(u \right)}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2} \cos{\left(u \right)} + 8 \cos{\left(u \right)}}{u}\, du = - \int \frac{u^{2} \cos{\left(u \right)} + 8 \cos{\left(u \right)}}{u}\, du$
      
      Rewrite the integrand:
      
      $\frac{u^{2} \cos{\left(u \right)} + 8 \cos{\left(u \right)}}{u} = u \cos{\left(u \right)} + \frac{8 \cos{\left(u \right)}}{u}$
      
      Integrate term-by-term:
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = \cos{\left(u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      Now evaluate the sub-integral.
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{8 \cos{\left(u \right)}}{u}\, du = 8 \int \frac{\cos{\left(u \right)}}{u}\, du$
      
      CiRule(a=1, b=0, context=cos(_u)/_u, symbol=_u)
      
      So, the result is: $8 \operatorname{Ci}{\left(u \right)}$
      
      The result is: $u \sin{\left(u \right)} + \cos{\left(u \right)} + 8 \operatorname{Ci}{\left(u \right)}$
      
      So, the result is: $- u \sin{\left(u \right)} - \cos{\left(u \right)} - 8 \operatorname{Ci}{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \cos{\left(\frac{1}{u} \right)} - 8 \operatorname{Ci}{\left(\frac{1}{u} \right)} - \frac{\sin{\left(\frac{1}{u} \right)}}{u}$
    So, the result is: $\cos{\left(\frac{1}{u} \right)} + 8 \operatorname{Ci}{\left(\frac{1}{u} \right)} + \frac{\sin{\left(\frac{1}{u} \right)}}{u}$
  Now substitute $u$ back in:
  
  $z \sin{\left(z \right)} + \cos{\left(z \right)} + 8 \operatorname{Ci}{\left(z \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\cos{\left(z \right)}}{z} \left(z^{2} + 8\right) = \frac{z^{2} \cos{\left(z \right)} + 8 \cos{\left(z \right)}}{z}$
2. Let $u = \frac{1}{z}$ .
  
  Then let $du = - \frac{dz}{z^{2}}$ and substitute $- du$ :
  
  $\int \left(- \frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}}\, du = - \int \frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}}\, du$
    1. Rewrite the integrand:
      
      $\frac{8 u^{2} \cos{\left(\frac{1}{u} \right)} + \cos{\left(\frac{1}{u} \right)}}{u^{3}} = \frac{8 \cos{\left(\frac{1}{u} \right)}}{u} + \frac{\cos{\left(\frac{1}{u} \right)}}{u^{3}}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{8 \cos{\left(\frac{1}{u} \right)}}{u}\, du = 8 \int \frac{\cos{\left(\frac{1}{u} \right)}}{u}\, du$
      
      Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\cos{\left(u \right)}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{u}\, du = - \int \frac{\cos{\left(u \right)}}{u}\, du$
      
      CiRule(a=1, b=0, context=cos(_u)/_u, symbol=_u)
      
      So, the result is: $- \operatorname{Ci}{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \operatorname{Ci}{\left(\frac{1}{u} \right)}$
      
      So, the result is: $- 8 \operatorname{Ci}{\left(\frac{1}{u} \right)}$
      
      Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- u \cos{\left(u \right)}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u \cos{\left(u \right)}\, du = - \int u \cos{\left(u \right)}\, du$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = \cos{\left(u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      Now evaluate the sub-integral.
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- u \sin{\left(u \right)} - \cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \cos{\left(\frac{1}{u} \right)} - \frac{\sin{\left(\frac{1}{u} \right)}}{u}$
      
      The result is: $- \cos{\left(\frac{1}{u} \right)} - 8 \operatorname{Ci}{\left(\frac{1}{u} \right)} - \frac{\sin{\left(\frac{1}{u} \right)}}{u}$
    So, the result is: $\cos{\left(\frac{1}{u} \right)} + 8 \operatorname{Ci}{\left(\frac{1}{u} \right)} + \frac{\sin{\left(\frac{1}{u} \right)}}{u}$
  Now substitute $u$ back in:
  
  $z \sin{\left(z \right)} + \cos{\left(z \right)} + 8 \operatorname{Ci}{\left(z \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\cos{\left(z \right)}}{z} \left(z^{2} + 8\right) = z \cos{\left(z \right)} + \frac{8 \cos{\left(z \right)}}{z}$
2. Integrate term-by-term:
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(z \right)} = z$ and let $\operatorname{dv}{\left(z \right)} = \cos{\left(z \right)}$ .
    
    Then $\operatorname{du}{\left(z \right)} = 1$ .
    
    To find $v{\left(z \right)}$ :
    1. The integral of cosine is sine:
      
      $\int \cos{\left(z \right)}\, dz = \sin{\left(z \right)}$
    Now evaluate the sub-integral.
  2. The integral of sine is negative cosine:
    
    $\int \sin{\left(z \right)}\, dz = - \cos{\left(z \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{8 \cos{\left(z \right)}}{z}\, dz = 8 \int \frac{\cos{\left(z \right)}}{z}\, dz$
    So, the result is: $8 \operatorname{Ci}{\left(z \right)}$
  The result is: $z \sin{\left(z \right)} + \cos{\left(z \right)} + 8 \operatorname{Ci}{\left(z \right)}$
Add the constant of integration:

$z \sin{\left(z \right)} + \cos{\left(z \right)} + 8 \operatorname{Ci}{\left(z \right)}+ \mathrm{constant}$

The answer is:

$z \sin{\left(z \right)} + \cos{\left(z \right)} + 8 \operatorname{Ci}{\left(z \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                    
 |                                                     
 | cos(z) / 2    \                                     
 | ------*\z  + 8/ dz = C + 8*Ci(z) + z*sin(z) + cos(z)
 |   z                                                 
 |                                                     
/

$$\int \frac{\cos{\left(z \right)}}{z} \left(z^{2} + 8\right)\, dz = C + z \sin{\left(z \right)} + \cos{\left(z \right)} + 8 \operatorname{Ci}{\left(z \right)}$$

Simplify

The answer [src]

-oo - cos(c) - 8*Ci(c) - c*sin(c)

$$- c \sin{\left(c \right)} - \cos{\left(c \right)} - 8 \operatorname{Ci}{\left(c \right)} - \infty$$

-oo - cos(c) - 8*Ci(c) - c*sin(c)

$$- c \sin{\left(c \right)} - \cos{\left(c \right)} - 8 \operatorname{Ci}{\left(c \right)} - \infty$$

-oo - cos(c) - 8*Ci(c) - c*sin(c)

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

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Identical expressions

Similar expressions

Integral of cosz/z(z^2+8) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3