Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of -e^x
Integral of -15x^2
Integral of xe^(1-x)
Integral of -xe^x
Identical expressions
tg(4x- two)
tg(4x minus 2)
tg(4x minus two)
tg4x-2
tg(4x-2)dx
Similar expressions
tg(4x+2)

Integral of tg(4x-2) dx

The solution

You have entered [src]

  3                
  /                
 |                 
 |  tan(4*x - 2) dx
 |                 
/                  
1

$$\int\limits_{1}^{3} \tan{\left(4 x - 2 \right)}\, dx$$

Integral(tan(4*x - 2), (x, 1, 3))

Detail solution

Rewrite the integrand:

$\tan{\left(4 x - 2 \right)} = \frac{\sin{\left(4 x - 2 \right)}}{\cos{\left(4 x - 2 \right)}}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \cos{\left(4 x - 2 \right)}$ .
  
  Then let $du = - 4 \sin{\left(4 x - 2 \right)} dx$ and substitute $- \frac{du}{4}$ :
  
  $\int \left(- \frac{1}{4 u}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{u}\, du = - \frac{\int \frac{1}{u}\, du}{4}$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    So, the result is: $- \frac{\log{\left(u \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\cos{\left(4 x - 2 \right)} \right)}}{4}$
Method #2
1. Let $u = 4 x - 2$ .
  
  Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
  
  $\int \frac{\sin{\left(u \right)}}{4 \cos{\left(u \right)}}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{\cos{\left(u \right)}}\, du = \frac{\int \frac{\sin{\left(u \right)}}{\cos{\left(u \right)}}\, du}{4}$
    1. Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{u}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $- \log{\left(\cos{\left(u \right)} \right)}$
    So, the result is: $- \frac{\log{\left(\cos{\left(u \right)} \right)}}{4}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\cos{\left(4 x - 2 \right)} \right)}}{4}$
Now simplify:

$- \frac{\log{\left(\cos{\left(4 x - 2 \right)} \right)}}{4}$
Add the constant of integration:

$- \frac{\log{\left(\cos{\left(4 x - 2 \right)} \right)}}{4}+ \mathrm{constant}$

The answer is:

$- \frac{\log{\left(\cos{\left(4 x - 2 \right)} \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                       
 |                       log(cos(4*x - 2))
 | tan(4*x - 2) dx = C - -----------------
 |                               4        
/

$$\int \tan{\left(4 x - 2 \right)}\, dx = C - \frac{\log{\left(\cos{\left(4 x - 2 \right)} \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     /       2   \      /       2    \
  log\1 + tan (2)/   log\1 + tan (10)/
- ---------------- + -----------------
         8                   8

$$- \frac{\log{\left(1 + \tan^{2}{\left(2 \right)} \right)}}{8} + \frac{\log{\left(\tan^{2}{\left(10 \right)} + 1 \right)}}{8}$$

     /       2   \      /       2    \
  log\1 + tan (2)/   log\1 + tan (10)/
- ---------------- + -----------------
         8                   8

$$- \frac{\log{\left(1 + \tan^{2}{\left(2 \right)} \right)}}{8} + \frac{\log{\left(\tan^{2}{\left(10 \right)} + 1 \right)}}{8}$$

-log(1 + tan(2)^2)/8 + log(1 + tan(10)^2)/8

Numerical answer [src]

1.98789651502355

1.98789651502355

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of tg(4x-2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2