Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of sin30
Integral of 1/cosx*sin^3x
Integral of e^(arccos(x))
Integral of 2sin^3x*cosx
Identical expressions
one /cosx*sin^3x
1 divide by co sinus of e of x multiply by sinus of cubed x
one divide by co sinus of e of x multiply by sinus of cubed x
1/cosx*sin3x
1/cosx*sin³x
1/cosx*sin to the power of 3x
1/cosxsin^3x
1/cosxsin3x
1 divide by cosx*sin^3x
1/cosx*sin^3xdx

Solving integrals/ 1/cosx*sin^3x

Integral of 1/cosx*sin^3x dx

The solution

You have entered [src]

  1                    
  /                    
 |                     
 |      1       3      
 |  1*------*sin (x) dx
 |    cos(x)           
 |                     
/                      
0

$$\int\limits_{0}^{1} 1 \cdot \frac{1}{\cos{\left(x \right)}} \sin^{3}{\left(x \right)}\, dx$$

Integral(1*sin(x)^3/cos(x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$1 \cdot \frac{1}{\cos{\left(x \right)}} \sin^{3}{\left(x \right)} = \frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos{\left(x \right)}}$
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 $du$ :
  
  $\int \frac{u^{2} - 1}{u}\, du$
  1. Let $u = u^{2}$ .
    
    Then let $du = 2 u du$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u - 1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u - 1}{2 u}\, du = \frac{\int \frac{u - 1}{u}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u - 1}{u} = 1 - \frac{1}{u}$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      The result is: $u - \log{\left(u \right)}$
      
      So, the result is: $\frac{u}{2} - \frac{\log{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{u^{2}}{2} - \frac{\log{\left(u^{2} \right)}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos{\left(x \right)}} = \frac{- \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
  
  $\int \frac{u^{2} - 1}{u}\, du$
  1. Let $u = u^{2}$ .
    
    Then let $du = 2 u du$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u - 1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u - 1}{2 u}\, du = \frac{\int \frac{u - 1}{u}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u - 1}{u} = 1 - \frac{1}{u}$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
      
      The result is: $u - \log{\left(u \right)}$
      
      So, the result is: $\frac{u}{2} - \frac{\log{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{u^{2}}{2} - \frac{\log{\left(u^{2} \right)}}{2}$
  Now substitute $u$ back in:
  
  $- \frac{\log{\left(\cos^{2}{\left(x \right)} \right)}}{2} + \frac{\cos^{2}{\left(x \right)}}{2}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos{\left(x \right)}} = - \sin{\left(x \right)} \cos{\left(x \right)} + \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      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:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
    So, the result is: $\frac{\cos^{2}{\left(x \right)}}{2}$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \frac{1}{u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $- \log{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- \log{\left(\cos{\left(x \right)} \right)}$
  The result is: $- \log{\left(\cos{\left(x \right)} \right)} + \frac{\cos^{2}{\left(x \right)}}{2}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                                
 |                              2         /   2   \
 |     1       3             cos (x)   log\cos (x)/
 | 1*------*sin (x) dx = C + ------- - ------------
 |   cos(x)                     2           2      
 |                                                 
/

$$-{{\log \left(\sin ^2x-1\right)}\over{2}}-{{\sin ^2x}\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

$$-{{\log \left(1-\sin ^21\right)}\over{2}}-{{\sin ^21}\over{2}}$$

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

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

Simplify

Numerical answer [src]

0.261589761249229

0.261589761249229

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of 1/cosx*sin^3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3