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Integral of d{x}:
Integral of e^3
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Identical expressions
sin^4tcos^2t
sinus of to the power of 4t co sinus of e of squared t
sin4tcos2t
sin⁴tcos²t
sin to the power of 4tcos to the power of 2t

Solving integrals/ sin^4tcos^2t

Integral of sin^4tcos^2t dt

The solution

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 2*pi                  
   /                   
  |                    
  |     4       2      
  |  sin (t)*cos (t) dt
  |                    
 /                     
 0

$$\int\limits_{0}^{2 \pi} \sin^{4}{\left(t \right)} \cos^{2}{\left(t \right)}\, dt$$

Integral(sin(t)^4*cos(t)^2, (t, 0, 2*pi))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(t \right)} \cos^{2}{\left(t \right)} = \left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right)$
There are multiple ways to do this integral.
Method #1
1. Let $u = 2 t$ .
  
  Then let $du = 2 dt$ and substitute $du$ :
  
  $\int \left(\frac{\cos^{3}{\left(u \right)}}{16} - \frac{\cos^{2}{\left(u \right)}}{16} - \frac{\cos{\left(u \right)}}{16} + \frac{1}{16}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos^{3}{\left(u \right)}}{16}\, du = \frac{\int \cos^{3}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{3}{\left(u \right)} = \left(1 - \sin^{2}{\left(u \right)}\right) \cos{\left(u \right)}$
      
      Let $u = \sin{\left(u \right)}$ .
      
      Then let $du = \cos{\left(u \right)} du$ and substitute $du$ :
      
      $\int \left(1 - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $- \frac{u^{3}}{3}$
      
      The result is: $- \frac{u^{3}}{3} + u$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(u \right)}}{3} + \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin^{3}{\left(u \right)}}{48} + \frac{\sin{\left(u \right)}}{16}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos^{2}{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos^{2}{\left(u \right)}\, du}{16}$
      
      Rewrite the integrand:
      
      $\cos^{2}{\left(u \right)} = \frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 u \right)}}{2}\, du = \frac{\int \cos{\left(2 u \right)}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 u \right)}}{2}$
      
      So, the result is: $\frac{\sin{\left(2 u \right)}}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The result is: $\frac{u}{2} + \frac{\sin{\left(2 u \right)}}{4}$
      
      So, the result is: $- \frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(u \right)}}{16}\right)\, du = - \frac{\int \cos{\left(u \right)}\, du}{16}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- \frac{\sin{\left(u \right)}}{16}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{16}\, du = \frac{u}{16}$
    The result is: $\frac{u}{32} - \frac{\sin{\left(2 u \right)}}{64} - \frac{\sin^{3}{\left(u \right)}}{48}$
  Now substitute $u$ back in:
  
  $\frac{t}{16} - \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$
Method #2
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(2 t \right)}}{8} - \frac{\cos^{2}{\left(2 t \right)}}{8} - \frac{\cos{\left(2 t \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{3}{\left(2 t \right)}}{8}\, dt = \frac{\int \cos^{3}{\left(2 t \right)}\, dt}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(2 t \right)} = \left(1 - \sin^{2}{\left(2 t \right)}\right) \cos{\left(2 t \right)}$
    2. Let $u = \sin{\left(2 t \right)}$ .
      
      Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 t \right)}}{6} + \frac{\sin{\left(2 t \right)}}{2}$
    So, the result is: $- \frac{\sin^{3}{\left(2 t \right)}}{48} + \frac{\sin{\left(2 t \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos^{2}{\left(2 t \right)}\, dt}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(2 t \right)} = \frac{\cos{\left(4 t \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\int \cos{\left(4 t \right)}\, dt}{2}$
      
      Let $u = 4 t$ .
      
      Then let $du = 4 dt$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 t \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 t \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dt = \frac{t}{2}$
      
      The result is: $\frac{t}{2} + \frac{\sin{\left(4 t \right)}}{8}$
    So, the result is: $- \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos{\left(2 t \right)}\, dt}{8}$
    1. Let $u = 2 t$ .
      
      Then let $du = 2 dt$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 t \right)}}{2}$
    So, the result is: $- \frac{\sin{\left(2 t \right)}}{16}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dt = \frac{t}{8}$
  The result is: $\frac{t}{16} - \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$
Method #3
1. Rewrite the integrand:
  
  $\left(\frac{1}{2} - \frac{\cos{\left(2 t \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 t \right)}}{2} + \frac{1}{2}\right) = \frac{\cos^{3}{\left(2 t \right)}}{8} - \frac{\cos^{2}{\left(2 t \right)}}{8} - \frac{\cos{\left(2 t \right)}}{8} + \frac{1}{8}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\cos^{3}{\left(2 t \right)}}{8}\, dt = \frac{\int \cos^{3}{\left(2 t \right)}\, dt}{8}$
    1. Rewrite the integrand:
      
      $\cos^{3}{\left(2 t \right)} = \left(1 - \sin^{2}{\left(2 t \right)}\right) \cos{\left(2 t \right)}$
    2. Let $u = \sin{\left(2 t \right)}$ .
      
      Then let $du = 2 \cos{\left(2 t \right)} dt$ and substitute $du$ :
      
      $\int \left(\frac{1}{2} - \frac{u^{2}}{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, du = \frac{u}{2}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{2}}{2}\right)\, du = - \frac{\int u^{2}\, du}{2}$
      
      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: $- \frac{u^{3}}{6} + \frac{u}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\sin^{3}{\left(2 t \right)}}{6} + \frac{\sin{\left(2 t \right)}}{2}$
    So, the result is: $- \frac{\sin^{3}{\left(2 t \right)}}{48} + \frac{\sin{\left(2 t \right)}}{16}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos^{2}{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos^{2}{\left(2 t \right)}\, dt}{8}$
    1. Rewrite the integrand:
      
      $\cos^{2}{\left(2 t \right)} = \frac{\cos{\left(4 t \right)}}{2} + \frac{1}{2}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 t \right)}}{2}\, dt = \frac{\int \cos{\left(4 t \right)}\, dt}{2}$
      
      Let $u = 4 t$ .
      
      Then let $du = 4 dt$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 t \right)}}{4}$
      
      So, the result is: $\frac{\sin{\left(4 t \right)}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dt = \frac{t}{2}$
      
      The result is: $\frac{t}{2} + \frac{\sin{\left(4 t \right)}}{8}$
    So, the result is: $- \frac{t}{16} - \frac{\sin{\left(4 t \right)}}{64}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\cos{\left(2 t \right)}}{8}\right)\, dt = - \frac{\int \cos{\left(2 t \right)}\, dt}{8}$
    1. Let $u = 2 t$ .
      
      Then let $du = 2 dt$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(2 t \right)}}{2}$
    So, the result is: $- \frac{\sin{\left(2 t \right)}}{16}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int \frac{1}{8}\, dt = \frac{t}{8}$
  The result is: $\frac{t}{16} - \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$
Add the constant of integration:

$\frac{t}{16} - \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}+ \mathrm{constant}$

The answer is:

$\frac{t}{16} - \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                  
 |                             3                     
 |    4       2             sin (2*t)   sin(4*t)   t 
 | sin (t)*cos (t) dt = C - --------- - -------- + --
 |                              48         64      16
/

$$\int \sin^{4}{\left(t \right)} \cos^{2}{\left(t \right)}\, dt = C + \frac{t}{16} - \frac{\sin^{3}{\left(2 t \right)}}{48} - \frac{\sin{\left(4 t \right)}}{64}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

pi
--
8

$$\frac{\pi}{8}$$

pi
--
8

$$\frac{\pi}{8}$$

pi/8

Numerical answer [src]

0.392699081698724

0.392699081698724

The graph

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

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Identical expressions

Integral of sin^4tcos^2t dt

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3