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Integral of sin^2xcos^5xdx
Identical expressions
sin^2xcos^5xdx
sinus of squared x co sinus of e of to the power of 5xdx
sin2xcos5xdx
sin²xcos⁵xdx
sin to the power of 2xcos to the power of 5xdx

Solving integrals/ sin^2xcos^5xdx

Integral of sin^2xcos^5xdx dx

The solution

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  1                     
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 |     2       5        
 |  sin (x)*cos (x)*1 dx
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0

$$\int\limits_{0}^{1} \sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} 1\, dx$$

Integral(sin(x)^2*cos(x)^5*1, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{2}{\left(x \right)} \cos^{5}{\left(x \right)} 1 = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{2}{\left(x \right)} \cos{\left(x \right)}$
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 $du$ :
  
  $\int \left(u^{6} - 2 u^{4} + u^{2}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 u^{4}\right)\, du = - 2 \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{2 u^{5}}{5}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
    The result is: $\frac{u^{7}}{7} - \frac{2 u^{5}}{5} + \frac{u^{3}}{3}$
  Now substitute $u$ back in:
  
  $\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{2}{\left(x \right)} \cos{\left(x \right)} = \sin^{6}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)} + \sin^{2}{\left(x \right)} \cos{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{6}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{7}{\left(x \right)}}{7}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
    So, the result is: $- \frac{2 \sin^{5}{\left(x \right)}}{5}$
  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:
    
    $\frac{\sin^{3}{\left(x \right)}}{3}$
  The result is: $\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{2}{\left(x \right)} \cos{\left(x \right)} = \sin^{6}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)} + \sin^{2}{\left(x \right)} \cos{\left(x \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(x \right)}$ .
    
    Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
    
    $\int u^{6}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{7}{\left(x \right)}}{7}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin^{4}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{4}{\left(x \right)} \cos{\left(x \right)}\, dx$
    1. Let $u = \sin{\left(x \right)}$ .
      
      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$ :
      
      $\int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(x \right)}}{5}$
    So, the result is: $- \frac{2 \sin^{5}{\left(x \right)}}{5}$
  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:
    
    $\frac{\sin^{3}{\left(x \right)}}{3}$
  The result is: $\frac{\sin^{7}{\left(x \right)}}{7} - \frac{2 \sin^{5}{\left(x \right)}}{5} + \frac{\sin^{3}{\left(x \right)}}{3}$
Now simplify:

$\frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}$
Add the constant of integration:

$\frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}+ \mathrm{constant}$

The answer is:

$\frac{\left(15 \sin^{4}{\left(x \right)} - 42 \sin^{2}{\left(x \right)} + 35\right) \sin^{3}{\left(x \right)}}{105}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                        
 |                                 5         3         7   
 |    2       5               2*sin (x)   sin (x)   sin (x)
 | sin (x)*cos (x)*1 dx = C - --------- + ------- + -------
 |                                5          3         7   
/

$${{15\,\sin ^7x-42\,\sin ^5x+35\,\sin ^3x}\over{105}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       5         3         7   
  2*sin (1)   sin (1)   sin (1)
- --------- + ------- + -------
      5          3         7

$${{15\,\sin ^71-42\,\sin ^51+35\,\sin ^31}\over{105}}$$

       5         3         7   
  2*sin (1)   sin (1)   sin (1)
- --------- + ------- + -------
      5          3         7

$$- \frac{2 \sin^{5}{\left(1 \right)}}{5} + \frac{\sin^{7}{\left(1 \right)}}{7} + \frac{\sin^{3}{\left(1 \right)}}{3}$$

Упростить

Numerical answer [src]

0.0725283477775366

0.0725283477775366

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of sin^2xcos^5xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3