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Solving integrals/ sinxcos^2(3x)

Integral of sinxcos^2(3x) dx

The solution

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  1                    
  /                    
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 |            2        
 |  sin(x)*cos (3*x) dx
 |                     
/                      
0

$$\int\limits_{0}^{1} \sin{\left(x \right)} \cos^{2}{\left(3 x \right)}\, dx$$

Integral(sin(x)*cos(3*x)^2, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin{\left(x \right)} \cos^{2}{\left(3 x \right)} = 16 \sin{\left(x \right)} \cos^{6}{\left(x \right)} - 24 \sin{\left(x \right)} \cos^{4}{\left(x \right)} + 9 \sin{\left(x \right)} \cos^{2}{\left(x \right)}$
Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 16 \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx = 16 \int \sin{\left(x \right)} \cos^{6}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{6}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{6}\right)\, du = - \int u^{6}\, du$
      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
        
        $\int u^{6}\, du = \frac{u^{7}}{7}$
      So, the result is: $- \frac{u^{7}}{7}$
    Now substitute $u$ back in:
    
    $- \frac{\cos^{7}{\left(x \right)}}{7}$
  So, the result is: $- \frac{16 \cos^{7}{\left(x \right)}}{7}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 24 \sin{\left(x \right)} \cos^{4}{\left(x \right)}\right)\, dx = - 24 \int \sin{\left(x \right)} \cos^{4}{\left(x \right)}\, dx$
  1. Let $u = \cos{\left(x \right)}$ .
    
    Then let $du = - \sin{\left(x \right)} dx$ and substitute $- du$ :
    
    $\int u^{4}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{4}\right)\, du = - \int u^{4}\, du$
      1. 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: $\frac{24 \cos^{5}{\left(x \right)}}{5}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int 9 \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = 9 \int \sin{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$
  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: $- 3 \cos^{3}{\left(x \right)}$
The result is: $- \frac{16 \cos^{7}{\left(x \right)}}{7} + \frac{24 \cos^{5}{\left(x \right)}}{5} - 3 \cos^{3}{\left(x \right)}$
Now simplify:

$- \frac{\left(80 \sin^{4}{\left(x \right)} + 8 \sin^{2}{\left(x \right)} + 17\right) \cos^{3}{\left(x \right)}}{35}$
Add the constant of integration:

$- \frac{\left(80 \sin^{4}{\left(x \right)} + 8 \sin^{2}{\left(x \right)} + 17\right) \cos^{3}{\left(x \right)}}{35}+ \mathrm{constant}$

The answer is:

$- \frac{\left(80 \sin^{4}{\left(x \right)} + 8 \sin^{2}{\left(x \right)} + 17\right) \cos^{3}{\left(x \right)}}{35}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                             
 |                                             7            5   
 |           2                    3      16*cos (x)   24*cos (x)
 | sin(x)*cos (3*x) dx = C - 3*cos (x) - ---------- + ----------
 |                                           7            5     
/

$$-{{5\,\cos \left(7\,x\right)-7\,\cos \left(5\,x\right)+70\,\cos x }\over{140}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

           2                   2                                   
17   18*sin (3)*cos(1)   17*cos (3)*cos(1)   6*cos(3)*sin(1)*sin(3)
-- - ----------------- - ----------------- + ----------------------
35           35                  35                    35

$${{17}\over{35}}-{{5\,\cos 7-7\,\cos 5+70\,\cos 1}\over{140}}$$

           2                   2                                   
17   18*sin (3)*cos(1)   17*cos (3)*cos(1)   6*cos(3)*sin(1)*sin(3)
-- - ----------------- - ----------------- + ----------------------
35           35                  35                    35

$$- \frac{17 \cos{\left(1 \right)} \cos^{2}{\left(3 \right)}}{35} + \frac{6 \sin{\left(1 \right)} \sin{\left(3 \right)} \cos{\left(3 \right)}}{35} - \frac{18 \sin^{2}{\left(3 \right)} \cos{\left(1 \right)}}{35} + \frac{17}{35}$$

Упростить

Numerical answer [src]

0.202821161541116

0.202821161541116

The graph

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

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Identical expressions

Integral of sinxcos^2(3x) dx

Limits of integration:

The graph:

Piecewise:

The solution