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Integral of d{x}:
Integral of log(sin(x))
Integral of cos(3x^2)
Integral of cos^5(2t)sin^2(2t)dt
Integral of xcosx/sin^2x
Identical expressions
cos^ five (two t)sin^2(2t)dt
co sinus of e of to the power of 5(2t) sinus of squared (2t)dt
co sinus of e of to the power of five (two t) sinus of squared (2t)dt
cos5(2t)sin2(2t)dt
cos52tsin22tdt
cos⁵(2t)sin²(2t)dt
cos to the power of 5(2t)sin to the power of 2(2t)dt
cos^52tsin^22tdt

Integral of cos^5(2t)sin^2(2t)dt dx

The solution

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  1                       
  /                       
 |                        
 |     5         2        
 |  cos (2*t)*sin (2*t) dt
 |                        
/                         
0

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

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

Detail solution

Rewrite the integrand:

$\sin^{2}{\left(2 t \right)} \cos^{5}{\left(2 t \right)} = \left(1 - \sin^{2}{\left(2 t \right)}\right)^{2} \sin^{2}{\left(2 t \right)} \cos{\left(2 t \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \sin{\left(2 t \right)}$ .
  
  Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $du$ :
  
  $\int \left(\frac{u^{6}}{2} - u^{4} + \frac{u^{2}}{2}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{6}}{2}\, du = \frac{\int u^{6}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $\frac{u^{7}}{14}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      So, the result is: $- \frac{u^{5}}{5}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
    The result is: $\frac{u^{7}}{14} - \frac{u^{5}}{5} + \frac{u^{3}}{6}$
  Now substitute $u$ back in:
  
  $\frac{\sin^{7}{\left(2 t \right)}}{14} - \frac{\sin^{5}{\left(2 t \right)}}{5} + \frac{\sin^{3}{\left(2 t \right)}}{6}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(2 t \right)}\right)^{2} \sin^{2}{\left(2 t \right)} \cos{\left(2 t \right)} = \sin^{6}{\left(2 t \right)} \cos{\left(2 t \right)} - 2 \sin^{4}{\left(2 t \right)} \cos{\left(2 t \right)} + \sin^{2}{\left(2 t \right)} \cos{\left(2 t \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(2 t \right)}$ .
    
    Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u^{6}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = \frac{\int u^{6}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $\frac{u^{7}}{14}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{7}{\left(2 t \right)}}{14}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin^{4}{\left(2 t \right)} \cos{\left(2 t \right)}\right)\, dt = - 2 \int \sin^{4}{\left(2 t \right)} \cos{\left(2 t \right)}\, dt$
    1. Let $u = \sin{\left(2 t \right)}$ .
      
      Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{4}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = \frac{\int u^{4}\, du}{2}$
      
      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: $\frac{u^{5}}{10}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(2 t \right)}}{10}$
    So, the result is: $- \frac{\sin^{5}{\left(2 t \right)}}{5}$
  1. Let $u = \sin{\left(2 t \right)}$ .
    
    Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u^{2}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{3}{\left(2 t \right)}}{6}$
  The result is: $\frac{\sin^{7}{\left(2 t \right)}}{14} - \frac{\sin^{5}{\left(2 t \right)}}{5} + \frac{\sin^{3}{\left(2 t \right)}}{6}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(2 t \right)}\right)^{2} \sin^{2}{\left(2 t \right)} \cos{\left(2 t \right)} = \sin^{6}{\left(2 t \right)} \cos{\left(2 t \right)} - 2 \sin^{4}{\left(2 t \right)} \cos{\left(2 t \right)} + \sin^{2}{\left(2 t \right)} \cos{\left(2 t \right)}$
2. Integrate term-by-term:
  1. Let $u = \sin{\left(2 t \right)}$ .
    
    Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u^{6}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = \frac{\int u^{6}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $\frac{u^{7}}{14}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{7}{\left(2 t \right)}}{14}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- 2 \sin^{4}{\left(2 t \right)} \cos{\left(2 t \right)}\right)\, dt = - 2 \int \sin^{4}{\left(2 t \right)} \cos{\left(2 t \right)}\, dt$
    1. Let $u = \sin{\left(2 t \right)}$ .
      
      Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{u^{4}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = \frac{\int u^{4}\, du}{2}$
      
      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: $\frac{u^{5}}{10}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{5}{\left(2 t \right)}}{10}$
    So, the result is: $- \frac{\sin^{5}{\left(2 t \right)}}{5}$
  1. Let $u = \sin{\left(2 t \right)}$ .
    
    Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{u^{2}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      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: $\frac{u^{3}}{6}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{3}{\left(2 t \right)}}{6}$
  The result is: $\frac{\sin^{7}{\left(2 t \right)}}{14} - \frac{\sin^{5}{\left(2 t \right)}}{5} + \frac{\sin^{3}{\left(2 t \right)}}{6}$
Add the constant of integration:

$\frac{\sin^{7}{\left(2 t \right)}}{14} - \frac{\sin^{5}{\left(2 t \right)}}{5} + \frac{\sin^{3}{\left(2 t \right)}}{6}+ \mathrm{constant}$

The answer is:

$\frac{\sin^{7}{\left(2 t \right)}}{14} - \frac{\sin^{5}{\left(2 t \right)}}{5} + \frac{\sin^{3}{\left(2 t \right)}}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                              
 |                                 5           3           7     
 |    5         2               sin (2*t)   sin (2*t)   sin (2*t)
 | cos (2*t)*sin (2*t) dt = C - --------- + --------- + ---------
 |                                  5           6           14   
/

$$\int \sin^{2}{\left(2 t \right)} \cos^{5}{\left(2 t \right)}\, dt = C + \frac{\sin^{7}{\left(2 t \right)}}{14} - \frac{\sin^{5}{\left(2 t \right)}}{5} + \frac{\sin^{3}{\left(2 t \right)}}{6}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     5         3         7   
  sin (2)   sin (2)   sin (2)
- ------- + ------- + -------
     5         6         14

$$- \frac{\sin^{5}{\left(2 \right)}}{5} + \frac{\sin^{7}{\left(2 \right)}}{14} + \frac{\sin^{3}{\left(2 \right)}}{6}$$

     5         3         7   
  sin (2)   sin (2)   sin (2)
- ------- + ------- + -------
     5         6         14

$$- \frac{\sin^{5}{\left(2 \right)}}{5} + \frac{\sin^{7}{\left(2 \right)}}{14} + \frac{\sin^{3}{\left(2 \right)}}{6}$$

-sin(2)^5/5 + sin(2)^3/6 + sin(2)^7/14

Numerical answer [src]

0.0376915856526081

0.0376915856526081

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

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How to use it?

Identical expressions

Integral of cos^5(2t)sin^2(2t)dt dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3