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(sinx-1)^2cosx

Integral of (sinx+1)^2cosx dx

The solution

You have entered [src]

 pi                        
 --                        
 6                         
  /                        
 |                         
 |              2          
 |  (sin(x) + 1) *cos(x) dx
 |                         
/                          
pi                         
--                         
2

$$\int\limits_{\frac{\pi}{2}}^{\frac{\pi}{6}} \left(\sin{\left(x \right)} + 1\right)^{2} \cos{\left(x \right)}\, dx$$

Integral((sin(x) + 1)^2*cos(x), (x, pi/2, pi/6))

Detail solution

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

$\frac{\left(\sin{\left(x \right)} + 1\right)^{3}}{3}$
Add the constant of integration:

$\frac{\left(\sin{\left(x \right)} + 1\right)^{3}}{3}+ \mathrm{constant}$

The answer is:

$\frac{\left(\sin{\left(x \right)} + 1\right)^{3}}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\int \left(\sin{\left(x \right)} + 1\right)^{2} \cos{\left(x \right)}\, dx = C + \frac{\left(\sin{\left(x \right)} + 1\right)^{3}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-37 
----
 24

$$- \frac{37}{24}$$

-37 
----
 24

$$- \frac{37}{24}$$

-37/24

Numerical answer [src]

-1.54166666666667

-1.54166666666667

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

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Identical expressions

Similar expressions

Integral of (sinx+1)^2cosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3