Integral of sec^4x dx

The solution

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  1           
  /           
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 |     4      
 |  sec (x) dx
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/             
0

$$\int\limits_{0}^{1} \sec^{4}{\left(x \right)}\, dx$$

Integral(sec(x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

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

$\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+ \mathrm{constant}$

The answer is:

$\frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                 
 |                     3            
 |    4             tan (x)         
 | sec (x) dx = C + ------- + tan(x)
 |                     3            
/

$$\int \sec^{4}{\left(x \right)}\, dx = C + \frac{\tan^{3}{\left(x \right)}}{3} + \tan{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  sin(1)    2*sin(1)
--------- + --------
     3      3*cos(1)
3*cos (1)

$$\frac{2 \sin{\left(1 \right)}}{3 \cos{\left(1 \right)}} + \frac{\sin{\left(1 \right)}}{3 \cos^{3}{\left(1 \right)}}$$

  sin(1)    2*sin(1)
--------- + --------
     3      3*cos(1)
3*cos (1)

$$\frac{2 \sin{\left(1 \right)}}{3 \cos{\left(1 \right)}} + \frac{\sin{\left(1 \right)}}{3 \cos^{3}{\left(1 \right)}}$$

sin(1)/(3*cos(1)^3) + 2*sin(1)/(3*cos(1))

Numerical answer [src]

2.81658164059915

2.81658164059915

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of sec^4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3