Integral of sin4xcosx dx

The solution

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$$\int\limits_{0}^{1} \sin{\left(4 x \right)} \cos{\left(x \right)}\, dx$$

Integral(sin(4*x)*cos(x), (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(4 x \right)} \cos{\left(x \right)} = - 8 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} + 4 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 8 \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\right)\, dx = - 8 \int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} = \left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
  2. There are multiple ways to do this integral.
    Method #1
    1. Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $du$ :
      
      $\int \left(u^{4} - u^{2}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{4}\, du = \frac{u^{5}}{5}$
      
      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: $\frac{u^{5}}{5} - \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
    Method #2
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{4}\, du$
      
      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: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{\cos^{5}{\left(x \right)}}{5}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
      
      The result is: $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
    Method #3
    1. Rewrite the integrand:
      
      $\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} = - \sin{\left(x \right)} \cos^{4}{\left(x \right)} + \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{4}\, du$
      
      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: $- \frac{u^{5}}{5}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{5}{\left(x \right)}}{5}$
      
      So, the result is: $\frac{\cos^{5}{\left(x \right)}}{5}$
      
      Let $u = \cos{\left(x \right)}$ .
      
      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
      
      $\int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{3}{\left(x \right)}}{3}$
      
      The result is: $\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $- \frac{8 \cos^{5}{\left(x \right)}}{5} + \frac{8 \cos^{3}{\left(x \right)}}{3}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 4 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 4 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{2}\, du = \frac{u^{3}}{3}$
      So, the result is: $- \frac{u^{3}}{3}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{3}{\left(x \right)}}{3}$
  So, the result is: $- \frac{4 \cos^{3}{\left(x \right)}}{3}$
The result is: $- \frac{8 \cos^{5}{\left(x \right)}}{5} + \frac{4 \cos^{3}{\left(x \right)}}{3}$
Now simplify:

$- \frac{\cos{\left(3 x \right)}}{6} - \frac{\cos{\left(5 x \right)}}{10}$
Add the constant of integration:

$- \frac{\cos{\left(3 x \right)}}{6} - \frac{\cos{\left(5 x \right)}}{10}+ \mathrm{constant}$

The answer is:

$- \frac{\cos{\left(3 x \right)}}{6} - \frac{\cos{\left(5 x \right)}}{10}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                              5           3   
 |                          8*cos (x)   4*cos (x)
 | sin(4*x)*cos(x) dx = C - --------- + ---------
 |                              5           3    
/

$$-{{\cos \left(5\,x\right)}\over{10}}-{{\cos \left(3\,x\right) }\over{6}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

4    4*cos(1)*cos(4)   sin(1)*sin(4)
-- - --------------- - -------------
15          15               15

$${{4}\over{15}}-{{3\,\cos 5+5\,\cos 3}\over{30}}$$

4    4*cos(1)*cos(4)   sin(1)*sin(4)
-- - --------------- - -------------
15          15               15

$$- \frac{\sin{\left(1 \right)} \sin{\left(4 \right)}}{15} - \frac{4 \cos{\left(1 \right)} \cos{\left(4 \right)}}{15} + \frac{4}{15}$$

Simplify

Numerical answer [src]

0.403299197553752

0.403299197553752

The graph

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

Other calculators

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Identical expressions

Similar expressions

Integral of sin4xcosx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3