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(sinx-2)^3cosxdx

Solving integrals/ (sinx+2)^3cosxdx

Integral of (sinx+2)^3cosxdx dx

The solution

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  1                          
  /                          
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 |              3            
 |  (sin(x) + 2) *cos(x)*1 dx
 |                           
/                            
0

$$\int\limits_{0}^{1} \left(\sin{\left(x \right)} + 2\right)^{3} \cos{\left(x \right)} 1\, dx$$

Integral((sin(x) + 2)^3*cos(x)*1, (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \sin{\left(x \right)} + 2$ .
  
  Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
  
  $\int u^{3}\, du$
  1. 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:
  
  $\frac{\left(\sin{\left(x \right)} + 2\right)^{4}}{4}$
Method #2
1. Rewrite the integrand:
  
  $\left(\sin{\left(x \right)} + 2\right)^{3} \cos{\left(x \right)} 1 = \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + 12 \sin{\left(x \right)} \cos{\left(x \right)} + 8 \cos{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{3}\, du$
    1. 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:
    
    $\frac{\sin^{4}{\left(x \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
    So, the result is: $2 \sin^{3}{\left(x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 12 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 12 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
    So, the result is: $- 6 \cos^{2}{\left(x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 \cos{\left(x \right)}\, dx = 8 \int \cos{\left(x \right)}\, dx$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
    So, the result is: $8 \sin{\left(x \right)}$
  The result is: $\frac{\sin^{4}{\left(x \right)}}{4} + 2 \sin^{3}{\left(x \right)} + 8 \sin{\left(x \right)} - 6 \cos^{2}{\left(x \right)}$
Method #3
1. Rewrite the integrand:
  
  $\left(\sin{\left(x \right)} + 2\right)^{3} \cos{\left(x \right)} 1 = \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 6 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + 12 \sin{\left(x \right)} \cos{\left(x \right)} + 8 \cos{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{3}\, du$
    1. 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:
    
    $\frac{\sin^{4}{\left(x \right)}}{4}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 6 \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx = 6 \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{3}{\left(x \right)}}{3}$
    So, the result is: $2 \sin^{3}{\left(x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 12 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 12 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(x \right)}}{2}$
    So, the result is: $- 6 \cos^{2}{\left(x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 8 \cos{\left(x \right)}\, dx = 8 \int \cos{\left(x \right)}\, dx$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$
    So, the result is: $8 \sin{\left(x \right)}$
  The result is: $\frac{\sin^{4}{\left(x \right)}}{4} + 2 \sin^{3}{\left(x \right)} + 8 \sin{\left(x \right)} - 6 \cos^{2}{\left(x \right)}$
Now simplify:

$\frac{\left(\sin{\left(x \right)} + 2\right)^{4}}{4}$
Add the constant of integration:

$\frac{\left(\sin{\left(x \right)} + 2\right)^{4}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\left(\sin{\left(x \right)} + 2\right)^{4}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                                             4
 |             3                   (sin(x) + 2) 
 | (sin(x) + 2) *cos(x)*1 dx = C + -------------
 |                                       4      
/

$${{\left(\sin x+2\right)^4}\over{4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                                      4   
     3           2                 sin (1)
2*sin (1) + 6*sin (1) + 8*sin(1) + -------
                                      4

$${{\sin ^41+8\,\sin ^31+24\,\sin ^21+32\,\sin 1+16}\over{4}}-4$$

                                      4   
     3           2                 sin (1)
2*sin (1) + 6*sin (1) + 8*sin(1) + -------
                                      4

$$\frac{\sin^{4}{\left(1 \right)}}{4} + 2 \sin^{3}{\left(1 \right)} + 6 \sin^{2}{\left(1 \right)} + 8 \sin{\left(1 \right)}$$

Simplify

Numerical answer [src]

12.2971968527029

12.2971968527029

The graph

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

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Identical expressions

Similar expressions

Integral of (sinx+2)^3cosxdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3