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(3sin^3x+6)cosx

Integral of (3sin^3x-6)cosx dx

The solution

You have entered [src]

  1                          
  /                          
 |                           
 |  /     3       \          
 |  \3*sin (x) - 6/*cos(x) dx
 |                           
/                            
0

$$\int\limits_{0}^{1} \left(3 \sin^{3}{\left(x \right)} - 6\right) \cos{\left(x \right)}\, dx$$

Integral((3*sin(x)^3 - 6)*cos(x), (x, 0, 1))

Detail solution

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

$\frac{3 \left(\sin^{3}{\left(x \right)} - 8\right) \sin{\left(x \right)}}{4}$
Add the constant of integration:

$\frac{3 \left(\sin^{3}{\left(x \right)} - 8\right) \sin{\left(x \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{3 \left(\sin^{3}{\left(x \right)} - 8\right) \sin{\left(x \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\int \left(3 \sin^{3}{\left(x \right)} - 6\right) \cos{\left(x \right)}\, dx = C + \frac{3 \sin^{4}{\left(x \right)}}{4} - 6 \sin{\left(x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                 4   
            3*sin (1)
-6*sin(1) + ---------
                4

$$- 6 \sin{\left(1 \right)} + \frac{3 \sin^{4}{\left(1 \right)}}{4}$$

                 4   
            3*sin (1)
-6*sin(1) + ---------
                4

$$- 6 \sin{\left(1 \right)} + \frac{3 \sin^{4}{\left(1 \right)}}{4}$$

-6*sin(1) + 3*sin(1)^4/4

Numerical answer [src]

-4.67279993459816

-4.67279993459816

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

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Similar expressions

Integral of (3sin^3x-6)cosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3