Integral of sin^7x dx

The solution

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 |     7      
 |  sin (x) dx
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0

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

Simplify

Detail solution

Rewrite the integrand:

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

$\frac{\cos^{7}{\left(x \right)}}{7} - \frac{3 \cos^{5}{\left(x \right)}}{5} + \cos^{3}{\left(x \right)} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer is:

$\frac{\cos^{7}{\left(x \right)}}{7} - \frac{3 \cos^{5}{\left(x \right)}}{5} + \cos^{3}{\left(x \right)} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                       
 |                                          5         7   
 |    7                3               3*cos (x)   cos (x)
 | sin (x) dx = C + cos (x) - cos(x) - --------- + -------
 |                                         5          7   
/

$${{\cos ^7x}\over{7}}-{{3\,\cos ^5x}\over{5}}+\cos ^3x-\cos x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                             5         7   
16      3               3*cos (1)   cos (1)
-- + cos (1) - cos(1) - --------- + -------
35                          5          7

$${{5\,\cos ^71-21\,\cos ^51+35\,\cos ^31-35\,\cos 1}\over{35}}+{{16 }\over{35}}$$

                             5         7   
16      3               3*cos (1)   cos (1)
-- + cos (1) - cos(1) - --------- + -------
35                          5          7

$$- \cos{\left(1 \right)} - \frac{3 \cos^{5}{\left(1 \right)}}{5} + \frac{\cos^{7}{\left(1 \right)}}{7} + \cos^{3}{\left(1 \right)} + \frac{16}{35}$$

Simplify

Numerical answer [src]

0.0488623115305527

0.0488623115305527

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 sin^7x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2