Integral of sin^6x*dx dx

The solution

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$$\int\limits_{0}^{1} \sin^{6}{\left(x \right)} 1\, dx$$

Integral(sin(x)^6*1, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{6}{\left(x \right)} 1 = \left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3}$
There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} = - \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
    2. There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)} = - \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{2}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      The result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
      
      Method #3
      
      Rewrite the integrand:
      
      $\left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)} = - \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)} + \cos{\left(2 x \right)}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\right)\, dx = - \int \sin^{2}{\left(2 x \right)} \cos{\left(2 x \right)}\, dx$
      
      Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{2}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      So, the result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6}$
      
      Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
      
      The result is: $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
    So, the result is: $\frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
    So, the result is: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 \cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
    So, the result is: $- \frac{3 \sin{\left(2 x \right)}}{16}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  The result is: $\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{3} = - \frac{\cos^{3}{\left(2 x \right)}}{8} + \frac{3 \cos^{2}{\left(2 x \right)}}{8} - \frac{3 \cos{\left(2 x \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{3}{\left(2 x \right)}}{8}\right)\, dx = - \frac{\int \cos^{3}{\left(2 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(2 x \right)} = \left(1 - \sin^{2}{\left(2 x \right)}\right) \cos{\left(2 x \right)}$
    2. Let $u = \sin{\left(2 x \right)}$ .
      
      Then let $du = 2 \cos{\left(2 x \right)} dx$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6}$
      
      The result is: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 x \right)}}{6} + \frac{\sin{\left(2 x \right)}}{2}$
    So, the result is: $\frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 \cos^{2}{\left(2 x \right)}}{8}\, dx = \frac{3 \int \cos^{2}{\left(2 x \right)}\, dx}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(2 x \right)} = \frac{\cos{\left(4 x \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 x \right)}}{2}\, dx = \frac{\int \cos{\left(4 x \right)}\, dx}{2}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 x \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
      
      The result is: $\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$
    So, the result is: $\frac{3 x}{16} + \frac{3 \sin{\left(4 x \right)}}{64}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{3 \cos{\left(2 x \right)}}{8}\right)\, dx = - \frac{3 \int \cos{\left(2 x \right)}\, dx}{8}$
    1. Let $u = 2 x$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 x \right)}}{2}$
    So, the result is: $- \frac{3 \sin{\left(2 x \right)}}{16}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dx = \frac{x}{8}$
  The result is: $\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}$
Add the constant of integration:

$\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}+ \mathrm{constant}$

The answer is:

$\frac{5 x}{16} + \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(2 x \right)}}{4} + \frac{3 \sin{\left(4 x \right)}}{64}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                          
 |                                  3                        
 |    6               sin(2*x)   sin (2*x)   3*sin(4*x)   5*x
 | sin (x)*1 dx = C - -------- + --------- + ---------- + ---
 |                       4           48          64        16
/

$${{{{3\,\left({{\sin \left(4\,x\right)}\over{2}}+2\,x\right)}\over{ 16}}-{{\sin \left(2\,x\right)-{{\sin ^3\left(2\,x\right)}\over{3}} }\over{8}}-{{3\,\sin \left(2\,x\right)}\over{8}}+{{x}\over{4}} }\over{2}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                            3                5          
5    5*cos(1)*sin(1)   5*sin (1)*cos(1)   sin (1)*cos(1)
-- - --------------- - ---------------- - --------------
16          16                24                6

$${{9\,\sin 4+4\,\sin ^32-48\,\sin 2+60}\over{192}}$$

                            3                5          
5    5*cos(1)*sin(1)   5*sin (1)*cos(1)   sin (1)*cos(1)
-- - --------------- - ---------------- - --------------
16          16                24                6

$$- \frac{5 \sin{\left(1 \right)} \cos{\left(1 \right)}}{16} - \frac{5 \sin^{3}{\left(1 \right)} \cos{\left(1 \right)}}{24} - \frac{\sin^{5}{\left(1 \right)} \cos{\left(1 \right)}}{6} + \frac{5}{16}$$

Simplify

Numerical answer [src]

0.0653635876732911

0.0653635876732911

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin^6x*dx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #3

Method #2