Integral of sin^3t dt

The solution

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 POST_GRBEK_SMALL_pi          
 -------------------          
          2                   
          /                   
         |                    
         |             3      
         |          sin (t) dt
         |                    
        /                     
        0

$$\int\limits_{0}^{\frac{POST_{GRBEK SMALL \pi}}{2}} \sin^{3}{\left(t \right)}\, dt$$

Integral(sin(t)^3, (t, 0, POST_GRBEK_SMALL_pi/2))

Detail solution

Rewrite the integrand:

$\sin^{3}{\left(t \right)} = \left(1 - \cos^{2}{\left(t \right)}\right) \sin{\left(t \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \cos{\left(t \right)}$ .
  
  Then let $du = - \sin{\left(t \right)} dt$ 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:
      
      $\int \left(-1\right)\, du = - u$
    The result is: $\frac{u^{3}}{3} - u$
  Now substitute $u$ back in:
  
  $\frac{\cos^{3}{\left(t \right)}}{3} - \cos{\left(t \right)}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(t \right)}\right) \sin{\left(t \right)} = - \sin{\left(t \right)} \cos^{2}{\left(t \right)} + \sin{\left(t \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(t \right)} \cos^{2}{\left(t \right)}\right)\, dt = - \int \sin{\left(t \right)} \cos^{2}{\left(t \right)}\, dt$
    1. Let $u = \cos{\left(t \right)}$ .
      
      Then let $du = - \sin{\left(t \right)} dt$ 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:
      
      $- \frac{\cos^{3}{\left(t \right)}}{3}$
    So, the result is: $\frac{\cos^{3}{\left(t \right)}}{3}$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(t \right)}\, dt = - \cos{\left(t \right)}$
  The result is: $\frac{\cos^{3}{\left(t \right)}}{3} - \cos{\left(t \right)}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \cos^{2}{\left(t \right)}\right) \sin{\left(t \right)} = - \sin{\left(t \right)} \cos^{2}{\left(t \right)} + \sin{\left(t \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin{\left(t \right)} \cos^{2}{\left(t \right)}\right)\, dt = - \int \sin{\left(t \right)} \cos^{2}{\left(t \right)}\, dt$
    1. Let $u = \cos{\left(t \right)}$ .
      
      Then let $du = - \sin{\left(t \right)} dt$ 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:
      
      $- \frac{\cos^{3}{\left(t \right)}}{3}$
    So, the result is: $\frac{\cos^{3}{\left(t \right)}}{3}$
  1. The integral of sine is negative cosine:
    
    $\int \sin{\left(t \right)}\, dt = - \cos{\left(t \right)}$
  The result is: $\frac{\cos^{3}{\left(t \right)}}{3} - \cos{\left(t \right)}$
Now simplify:

$\frac{\left(\cos{\left(2 t \right)} - 5\right) \cos{\left(t \right)}}{6}$
Add the constant of integration:

$\frac{\left(\cos{\left(2 t \right)} - 5\right) \cos{\left(t \right)}}{6}+ \mathrm{constant}$

The answer is:

$\frac{\left(\cos{\left(2 t \right)} - 5\right) \cos{\left(t \right)}}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                 
 |                              3   
 |    3                      cos (t)
 | sin (t) dt = C - cos(t) + -------
 |                              3   
/

$${{\cos ^3t}\over{3}}-\cos t$$

Simplify

The answer [src]

                                  3/POST_GRBEK_SMALL_pi\
                               cos |-------------------|
2      /POST_GRBEK_SMALL_pi\       \         2         /
- - cos|-------------------| + -------------------------
3      \         2         /               3

$$\frac{\cos^{3}{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)}}{3} - \cos{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)} + \frac{2}{3}$$

                                  3/POST_GRBEK_SMALL_pi\
                               cos |-------------------|
2      /POST_GRBEK_SMALL_pi\       \         2         /
- - cos|-------------------| + -------------------------
3      \         2         /               3

$$\frac{\cos^{3}{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)}}{3} - \cos{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)} + \frac{2}{3}$$

Simplify

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

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Identical expressions

Integral of sin^3t dt

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3