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Integral of d{x}:
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Integral of cos^3(3t)sin(3t)dt
Identical expressions
cos^ three (3t)sin(3t)dt
co sinus of e of cubed (3t) sinus of (3t)dt
co sinus of e of to the power of three (3t) sinus of (3t)dt
cos3(3t)sin(3t)dt
cos33tsin3tdt
cos³(3t)sin(3t)dt
cos to the power of 3(3t)sin(3t)dt
cos^33tsin3tdt

Solving integrals/ cos^3(3t)sin(3t)dt

Integral of cos^3(3t)sin(3t)dt dx

The solution

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  1                        
  /                        
 |                         
 |     3                   
 |  cos (3*t)*sin(3*t)*1 dt
 |                         
/                          
0

$$\int\limits_{0}^{1} \cos^{3}{\left(3 t \right)} \sin{\left(3 t \right)} 1\, dt$$

Integral(cos(3*t)^3*sin(3*t)*1, (t, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \cos{\left(3 t \right)}$ .
  
  Then let $du = - 3 \sin{\left(3 t \right)} dt$ and substitute $- \frac{du}{3}$ :
  
  $\int \frac{u^{3}}{9}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{u^{3}}{3}\right)\, du = - \frac{\int u^{3}\, du}{3}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
    So, the result is: $- \frac{u^{4}}{12}$
  Now substitute $u$ back in:
  
  $- \frac{\cos^{4}{\left(3 t \right)}}{12}$
Method #2
1. Rewrite the integrand:
  
  $\cos^{3}{\left(3 t \right)} \sin{\left(3 t \right)} 1 = \left(1 - \sin^{2}{\left(3 t \right)}\right) \sin{\left(3 t \right)} \cos{\left(3 t \right)}$
2. Let $u = \sin^{2}{\left(3 t \right)}$ .
  
  Then let $du = 6 \sin{\left(3 t \right)} \cos{\left(3 t \right)} dt$ and substitute $du$ :
  
  $\int \left(\frac{1}{6} - \frac{u}{6}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{6}\, du = \frac{u}{6}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u}{6}\right)\, du = - \frac{\int u\, du}{6}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{12}$
    The result is: $- \frac{u^{2}}{12} + \frac{u}{6}$
  Now substitute $u$ back in:
  
  $- \frac{\sin^{4}{\left(3 t \right)}}{12} + \frac{\sin^{2}{\left(3 t \right)}}{6}$
Add the constant of integration:

$- \frac{\cos^{4}{\left(3 t \right)}}{12}+ \mathrm{constant}$

The answer is:

$- \frac{\cos^{4}{\left(3 t \right)}}{12}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                       
 |                                  4     
 |    3                          cos (3*t)
 | cos (3*t)*sin(3*t)*1 dt = C - ---------
 |                                   12   
/

$$-{{\cos ^4\left(3\,t\right)}\over{12}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

        4   
1    cos (3)
-- - -------
12      12

$${{1}\over{12}}-{{\cos ^43}\over{12}}$$

        4   
1    cos (3)
-- - -------
12      12

$$- \frac{\cos^{4}{\left(3 \right)}}{12} + \frac{1}{12}$$

Simplify

Numerical answer [src]

0.00328609265277129

0.00328609265277129

The graph

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

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Identical expressions

Integral of cos^3(3t)sin(3t)dt dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2