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Integral of d{x}:
Integral of sin^3x/cos^7x
Integral of 3x
Integral of tan^3(x)/cos^2(x)
Integral of cos(3x^2+4)
Identical expressions
sin^3x/cos^7x
sinus of cubed x divide by co sinus of e of to the power of 7x
sin3x/cos7x
sin³x/cos⁷x
sin to the power of 3x/cos to the power of 7x
sin^3x divide by cos^7x
sin^3x/cos^7xdx

Solving integrals/ sin^3x/cos^7x

Integral of sin^3x/cos^7x dx

The solution

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  1           
  /           
 |            
 |     3      
 |  sin (x)   
 |  ------- dx
 |     7      
 |  cos (x)   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{\sin^{3}{\left(x \right)}}{\cos^{7}{\left(x \right)}}\, dx$$

Simplify

Detail solution

Rewrite the integrand:

$\frac{\sin^{3}{\left(x \right)}}{\cos^{7}{\left(x \right)}} = \frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{7}{\left(x \right)}}$
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 $du$ :
  
  $\int \frac{u^{2} - 1}{u^{7}}\, du$
  1. Let $u = u^{2}$ .
    
    Then let $du = 2 u du$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u - 1}{4 u^{4}}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u - 1}{2 u^{4}}\, du = \frac{\int \frac{u - 1}{u^{4}}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u - 1}{u^{4}} = \frac{1}{u^{3}} - \frac{1}{u^{4}}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u^{4}}\right)\, du = - \int \frac{1}{u^{4}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$
      
      So, the result is: $\frac{1}{3 u^{3}}$
      
      The result is: $- \frac{1}{2 u^{2}} + \frac{1}{3 u^{3}}$
      
      So, the result is: $- \frac{1}{4 u^{2}} + \frac{1}{6 u^{3}}$
    Now substitute $u$ back in:
    
    $- \frac{1}{4 u^{4}} + \frac{1}{6 u^{6}}$
  Now substitute $u$ back in:
  
  $- \frac{1}{4 \cos^{4}{\left(x \right)}} + \frac{1}{6 \cos^{6}{\left(x \right)}}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{7}{\left(x \right)}} = \frac{- \sin{\left(x \right)} \cos^{2}{\left(x \right)} + \sin{\left(x \right)}}{\cos^{7}{\left(x \right)}}$
2. Let $u = \cos{\left(x \right)}$ .
  
  Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
  
  $\int \frac{u^{2} - 1}{u^{7}}\, du$
  1. Let $u = u^{2}$ .
    
    Then let $du = 2 u du$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u - 1}{4 u^{4}}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u - 1}{2 u^{4}}\, du = \frac{\int \frac{u - 1}{u^{4}}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u - 1}{u^{4}} = \frac{1}{u^{3}} - \frac{1}{u^{4}}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u^{4}}\right)\, du = - \int \frac{1}{u^{4}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$
      
      So, the result is: $\frac{1}{3 u^{3}}$
      
      The result is: $- \frac{1}{2 u^{2}} + \frac{1}{3 u^{3}}$
      
      So, the result is: $- \frac{1}{4 u^{2}} + \frac{1}{6 u^{3}}$
    Now substitute $u$ back in:
    
    $- \frac{1}{4 u^{4}} + \frac{1}{6 u^{6}}$
  Now substitute $u$ back in:
  
  $- \frac{1}{4 \cos^{4}{\left(x \right)}} + \frac{1}{6 \cos^{6}{\left(x \right)}}$
Method #3
1. Rewrite the integrand:
  
  $\frac{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)}}{\cos^{7}{\left(x \right)}} = - \frac{\sin{\left(x \right)}}{\cos^{5}{\left(x \right)}} + \frac{\sin{\left(x \right)}}{\cos^{7}{\left(x \right)}}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\sin{\left(x \right)}}{\cos^{5}{\left(x \right)}}\right)\, dx = - \int \frac{\sin{\left(x \right)}}{\cos^{5}{\left(x \right)}}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int \frac{1}{u^{5}}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u^{5}}\right)\, du = - \int \frac{1}{u^{5}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{5}}\, du = - \frac{1}{4 u^{4}}$
      
      So, the result is: $\frac{1}{4 u^{4}}$
      
      Now substitute $u$ back in:
      
      $\frac{1}{4 \cos^{4}{\left(x \right)}}$
    So, the result is: $- \frac{1}{4 \cos^{4}{\left(x \right)}}$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int \frac{1}{u^{7}}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u^{7}}\right)\, du = - \int \frac{1}{u^{7}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{7}}\, du = - \frac{1}{6 u^{6}}$
      
      So, the result is: $\frac{1}{6 u^{6}}$
    Now substitute $u$ back in:
    
    $\frac{1}{6 \cos^{6}{\left(x \right)}}$
  The result is: $- \frac{1}{4 \cos^{4}{\left(x \right)}} + \frac{1}{6 \cos^{6}{\left(x \right)}}$
Now simplify:

$\frac{2 - 3 \cos^{2}{\left(x \right)}}{12 \cos^{6}{\left(x \right)}}$
Add the constant of integration:

$\frac{2 - 3 \cos^{2}{\left(x \right)}}{12 \cos^{6}{\left(x \right)}}+ \mathrm{constant}$

The answer is:

$\frac{2 - 3 \cos^{2}{\left(x \right)}}{12 \cos^{6}{\left(x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                      
 |                                       
 |    3                                  
 | sin (x)              1           1    
 | ------- dx = C - --------- + ---------
 |    7                  4           6   
 | cos (x)          4*cos (x)   6*cos (x)
 |                                       
/

$$-{{3\,\sin ^2x-1}\over{12\,\sin ^6x-36\,\sin ^4x+36\,\sin ^2x-12}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

              2   
1    2 - 3*cos (1)
-- + -------------
12           6    
       12*cos (1)

$${{1}\over{12\,\sin ^61-36\,\sin ^41+36\,\sin ^21-12}}-{{\sin ^21 }\over{4\,\sin ^61-12\,\sin ^41+12\,\sin ^21-4}}+{{1}\over{12}}$$

              2   
1    2 - 3*cos (1)
-- + -------------
12           6    
       12*cos (1)

$$\frac{1}{12} + \frac{- 3 \cos^{2}{\left(1 \right)} + 2}{12 \cos^{6}{\left(1 \right)}}$$

Simplify

Numerical answer [src]

3.84906381342323

3.84906381342323

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin^3x/cos^7x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3