Integral of tgx(ln(cosx)) dx

The solution

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 |  tan(x)*log(cos(x)) dx
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$$\int\limits_{0}^{1} \log{\left(\cos{\left(x \right)} \right)} \tan{\left(x \right)}\, dx$$

Integral(tan(x)*log(cos(x)), (x, 0, 1))

Detail solution

Use integration by parts:

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

Let $u{\left(x \right)} = \log{\left(\cos{\left(x \right)} \right)}$ and let $\operatorname{dv}{\left(x \right)} = \tan{\left(x \right)}$ .

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $- \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $- \log{\left(\cos{\left(x \right)} \right)}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \log{\left(\cos{\left(x \right)} \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$ .

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $- \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $- \log{\left(\cos{\left(x \right)} \right)}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \log{\left(\cos{\left(x \right)} \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$ .

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $- \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $- \log{\left(\cos{\left(x \right)} \right)}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \log{\left(\cos{\left(x \right)} \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$ .

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $- \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $- \log{\left(\cos{\left(x \right)} \right)}$
Now evaluate the sub-integral.
Use integration by parts:

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

Let $u{\left(x \right)} = \log{\left(\cos{\left(x \right)} \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$ .

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

To find $v{\left(x \right)}$ :
1. Rewrite the integrand:
  
  $\tan{\left(x \right)} = \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{1}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $- \log{\left(u \right)}$
  Now substitute $u$ back in:
  
  $- \log{\left(\cos{\left(x \right)} \right)}$
Now evaluate the sub-integral.
There are multiple ways to do this integral.
Method #1
1. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
  
  $\int \left(- \frac{\log{\left(u \right)}}{u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\log{\left(u \right)}}{u}\, du = - \int \frac{\log{\left(u \right)}}{u}\, du$
    1. Let $u = \frac{1}{u}$ .
      
      Then let $du = - \frac{du}{u^{2}}$ and substitute $- du$ :
      
      $\int \left(- \frac{\log{\left(\frac{1}{u} \right)}}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du$
      
      Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $- du$ :
      
      $\int \left(- u\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$
      
      So, the result is: $\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(u \right)}^{2}}{2}$
    So, the result is: $- \frac{\log{\left(u \right)}^{2}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\cos{\left(x \right)} \right)}^{2}}{2}$
Method #2
1. Let $u = \log{\left(\cos{\left(x \right)} \right)}$ .
  
  Then let $du = - \frac{\sin{\left(x \right)} dx}{\cos{\left(x \right)}}$ and substitute $- du$ :
  
  $\int \left(- u\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u\, du = - \int u\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
    So, the result is: $- \frac{u^{2}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\cos{\left(x \right)} \right)}^{2}}{2}$
Add the constant of integration:

$- \frac{\log{\left(\cos{\left(x \right)} \right)}^{2}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(\cos{\left(x \right)} \right)}^{2}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                               2        
 |                             log (cos(x))
 | tan(x)*log(cos(x)) dx = C - ------------
 |                                  2      
/

$$\int \log{\left(\cos{\left(x \right)} \right)} \tan{\left(x \right)}\, dx = C - \frac{\log{\left(\cos{\left(x \right)} \right)}^{2}}{2}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

    2         
-log (cos(1)) 
--------------
      2

$$- \frac{\log{\left(\cos{\left(1 \right)} \right)}^{2}}{2}$$

    2         
-log (cos(1)) 
--------------
      2

$$- \frac{\log{\left(\cos{\left(1 \right)} \right)}^{2}}{2}$$

-log(cos(1))^2/2

Numerical answer [src]

-0.189497975519971

-0.189497975519971

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 tgx(ln(cosx)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2