Integral of dz÷(5z+tan(y-3x)) dx

The solution

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  1                        
  /                        
 |                         
 |            1            
 |  1*------------------ dz
 |    5*z + tan(y - 3*x)   
 |                         
/                          
0

$$\int\limits_{0}^{1} 1 \cdot \frac{1}{5 z + \tan{\left(- 3 x + y \right)}}\, dz$$

Integral(1/(5*z + tan(y - 3*x)), (z, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 5 z + \tan{\left(- 3 x + y \right)}$ .
  
  Then let $du = 5 dz$ and substitute $\frac{du}{5}$ :
  
  $\int \frac{1}{25 u}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{5 u}\, du = \frac{\int \frac{1}{u}\, du}{5}$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $\frac{\log{\left(u \right)}}{5}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(5 z + \tan{\left(- 3 x + y \right)} \right)}}{5}$
Method #2
1. Rewrite the integrand:
  
  $1 \cdot \frac{1}{5 z + \tan{\left(- 3 x + y \right)}} = \frac{1}{5 z - \tan{\left(3 x - y \right)}}$
2. Let $u = 5 z - \tan{\left(3 x - y \right)}$ .
  
  Then let $du = 5 dz$ and substitute $\frac{du}{5}$ :
  
  $\int \frac{1}{25 u}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{5 u}\, du = \frac{\int \frac{1}{u}\, du}{5}$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $\frac{\log{\left(u \right)}}{5}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(5 z - \tan{\left(3 x - y \right)} \right)}}{5}$
Method #3
1. Rewrite the integrand:
  
  $1 \cdot \frac{1}{5 z + \tan{\left(- 3 x + y \right)}} = \frac{1}{5 z - \tan{\left(3 x - y \right)}}$
2. Let $u = 5 z - \tan{\left(3 x - y \right)}$ .
  
  Then let $du = 5 dz$ and substitute $\frac{du}{5}$ :
  
  $\int \frac{1}{25 u}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{5 u}\, du = \frac{\int \frac{1}{u}\, du}{5}$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $\frac{\log{\left(u \right)}}{5}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(5 z - \tan{\left(3 x - y \right)} \right)}}{5}$
Now simplify:

$\frac{\log{\left(5 z - \tan{\left(3 x - y \right)} \right)}}{5}$
Add the constant of integration:

$\frac{\log{\left(5 z - \tan{\left(3 x - y \right)} \right)}}{5}+ \mathrm{constant}$

The answer is:

$\frac{\log{\left(5 z - \tan{\left(3 x - y \right)} \right)}}{5}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     
 |                                                      
 |           1                   log(5*z + tan(y - 3*x))
 | 1*------------------ dz = C + -----------------------
 |   5*z + tan(y - 3*x)                     5           
 |                                                      
/

$${{\log \left(5\,z+\tan \left(y-3\,x\right)\right)}\over{5}}$$

Simplify

The answer [src]

  log(-tan(-y + 3*x))   log(5 - tan(-y + 3*x))
- ------------------- + ----------------------
           5                      5

$$\frac{\log{\left(- \tan{\left(3 x - y \right)} + 5 \right)}}{5} - \frac{\log{\left(- \tan{\left(3 x - y \right)} \right)}}{5}$$

  log(-tan(-y + 3*x))   log(5 - tan(-y + 3*x))
- ------------------- + ----------------------
           5                      5

$$\frac{\log{\left(- \tan{\left(3 x - y \right)} + 5 \right)}}{5} - \frac{\log{\left(- \tan{\left(3 x - y \right)} \right)}}{5}$$

Simplify

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

Other calculators

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Identical expressions

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Integral of dz÷(5z+tan(y-3x)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3