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1/tgx*cos^2xdx

Integral of 1/tgx*cos^2x dx

The solution

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  1           
  /           
 |            
 |     2      
 |  cos (x)   
 |  ------- dx
 |   tan(x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\cos^{2}{\left(x \right)}}{\tan{\left(x \right)}}\, dx$$

Integral(cos(x)^2/tan(x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\frac{1}{\tan{\left(x \right)} \sec^{2}{\left(x \right)}} = \frac{\tan{\left(x \right)}}{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{2}{\left(x \right)}}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \sec^{2}{\left(x \right)}$ .
  
  Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $du$ :
  
  $\int \frac{1}{2 u^{3} - 2 u^{2}}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{2 u^{3} - 2 u^{2}} = \frac{1}{2 \left(u - 1\right)} - \frac{1}{2 u} - \frac{1}{2 u^{2}}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(u - 1\right)}\, du = \frac{\int \frac{1}{u - 1}\, du}{2}$
      
      Let $u = u - 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 1 \right)}$
      
      So, the result is: $\frac{\log{\left(u - 1 \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 u}\right)\, du = - \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \frac{\log{\left(u \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 u^{2}}\right)\, du = - \frac{\int \frac{1}{u^{2}}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      So, the result is: $\frac{1}{2 u}$
    The result is: $- \frac{\log{\left(u \right)}}{2} + \frac{\log{\left(u - 1 \right)}}{2} + \frac{1}{2 u}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(\sec^{2}{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{1}{2 \sec^{2}{\left(x \right)}}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\tan{\left(x \right)}}{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{2}{\left(x \right)}} = \frac{\tan{\left(x \right)}}{\sec^{4}{\left(x \right)} - \sec^{2}{\left(x \right)}}$
2. Let $u = \sec^{2}{\left(x \right)}$ .
  
  Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $du$ :
  
  $\int \frac{1}{2 u^{3} - 2 u^{2}}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{2 u^{3} - 2 u^{2}} = \frac{1}{2 \left(u - 1\right)} - \frac{1}{2 u} - \frac{1}{2 u^{2}}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(u - 1\right)}\, du = \frac{\int \frac{1}{u - 1}\, du}{2}$
      
      Let $u = u - 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 1 \right)}$
      
      So, the result is: $\frac{\log{\left(u - 1 \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 u}\right)\, du = - \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \frac{\log{\left(u \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 u^{2}}\right)\, du = - \frac{\int \frac{1}{u^{2}}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      So, the result is: $\frac{1}{2 u}$
    The result is: $- \frac{\log{\left(u \right)}}{2} + \frac{\log{\left(u - 1 \right)}}{2} + \frac{1}{2 u}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(\sec^{2}{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{1}{2 \sec^{2}{\left(x \right)}}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\tan{\left(x \right)}}{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{2}{\left(x \right)}} = \frac{\tan{\left(x \right)}}{\sec^{4}{\left(x \right)} - \sec^{2}{\left(x \right)}}$
2. Let $u = \sec^{2}{\left(x \right)}$ .
  
  Then let $du = 2 \tan{\left(x \right)} \sec^{2}{\left(x \right)} dx$ and substitute $du$ :
  
  $\int \frac{1}{2 u^{3} - 2 u^{2}}\, du$
  1. Rewrite the integrand:
    
    $\frac{1}{2 u^{3} - 2 u^{2}} = \frac{1}{2 \left(u - 1\right)} - \frac{1}{2 u} - \frac{1}{2 u^{2}}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 \left(u - 1\right)}\, du = \frac{\int \frac{1}{u - 1}\, du}{2}$
      
      Let $u = u - 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u - 1 \right)}$
      
      So, the result is: $\frac{\log{\left(u - 1 \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 u}\right)\, du = - \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \frac{\log{\left(u \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{2 u^{2}}\right)\, du = - \frac{\int \frac{1}{u^{2}}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$
      
      So, the result is: $\frac{1}{2 u}$
    The result is: $- \frac{\log{\left(u \right)}}{2} + \frac{\log{\left(u - 1 \right)}}{2} + \frac{1}{2 u}$
  Now substitute $u$ back in:
  
  $\frac{\log{\left(\sec^{2}{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{1}{2 \sec^{2}{\left(x \right)}}$
Now simplify:

$- \frac{\log{\left(\frac{1}{\cos^{2}{\left(x \right)}} \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}$
Add the constant of integration:

$- \frac{\log{\left(\frac{1}{\cos^{2}{\left(x \right)}} \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(\frac{1}{\cos^{2}{\left(x \right)}} \right)}}{2} + \frac{\log{\left(\tan^{2}{\left(x \right)} \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                             
 |                                                              
 |    2                            /        2   \      /   2   \
 | cos (x)              1       log\-1 + sec (x)/   log\sec (x)/
 | ------- dx = C + --------- + ----------------- - ------------
 |  tan(x)               2              2                2      
 |                  2*sec (x)                                   
/

$$\int \frac{\cos^{2}{\left(x \right)}}{\tan{\left(x \right)}}\, dx = C + \frac{\log{\left(\sec^{2}{\left(x \right)} - 1 \right)}}{2} - \frac{\log{\left(\sec^{2}{\left(x \right)} \right)}}{2} + \frac{1}{2 \sec^{2}{\left(x \right)}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

$$\infty$$

oo

$$\infty$$

oo

Numerical answer [src]

43.563805678587

43.563805678587

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

Other calculators

How to use it?

Identical expressions

Integral of 1/tgx*cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3