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Identical expressions
cos^ three (five *x)*sin^ four (five *x)
co sinus of e of cubed (5 multiply by x) multiply by sinus of to the power of 4(5 multiply by x)
co sinus of e of to the power of three (five multiply by x) multiply by sinus of to the power of four (five multiply by x)
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cos35*x*sin45*x
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cos^3(5x)sin^4(5x)
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cos^3(5*x)*sin^4(5*x)dx

Solving integrals/ cos^3(5*x)*sin^4(5*x)

Integral of cos^3(5x)sin^4(5*x) dx

The solution

You have entered [src]

  1                       
  /                       
 |                        
 |     3         4        
 |  cos (5*x)*sin (5*x) dx
 |                        
/                         
0

$$\int\limits_{0}^{1} \sin^{4}{\left(5 x \right)} \cos^{3}{\left(5 x \right)}\, dx$$

Integral(cos(5*x)^3*sin(5*x)^4, (x, 0, 1))

Detail solution

Rewrite the integrand:

$\sin^{4}{\left(5 x \right)} \cos^{3}{\left(5 x \right)} = \left(1 - \sin^{2}{\left(5 x \right)}\right) \sin^{4}{\left(5 x \right)} \cos{\left(5 x \right)}$
There are multiple ways to do this integral.
Method #1
1. Let $u = \sin{\left(5 x \right)}$ .
  
  Then let $du = 5 \cos{\left(5 x \right)} dx$ and substitute $du$ :
  
  $\int \left(- \frac{u^{6}}{5} + \frac{u^{4}}{5}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u^{6}}{5}\right)\, du = - \frac{\int u^{6}\, du}{5}$
      
      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: $- \frac{u^{7}}{35}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{4}}{5}\, du = \frac{\int u^{4}\, du}{5}$
      
      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: $\frac{u^{5}}{25}$
    The result is: $- \frac{u^{7}}{35} + \frac{u^{5}}{25}$
  Now substitute $u$ back in:
  
  $- \frac{\sin^{7}{\left(5 x \right)}}{35} + \frac{\sin^{5}{\left(5 x \right)}}{25}$
Method #2
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(5 x \right)}\right) \sin^{4}{\left(5 x \right)} \cos{\left(5 x \right)} = - \sin^{6}{\left(5 x \right)} \cos{\left(5 x \right)} + \sin^{4}{\left(5 x \right)} \cos{\left(5 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin^{6}{\left(5 x \right)} \cos{\left(5 x \right)}\right)\, dx = - \int \sin^{6}{\left(5 x \right)} \cos{\left(5 x \right)}\, dx$
    1. Let $u = \sin{\left(5 x \right)}$ .
      
      Then let $du = 5 \cos{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{u^{6}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = \frac{\int u^{6}\, du}{5}$
      
      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: $\frac{u^{7}}{35}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(5 x \right)}}{35}$
    So, the result is: $- \frac{\sin^{7}{\left(5 x \right)}}{35}$
  1. Let $u = \sin{\left(5 x \right)}$ .
    
    Then let $du = 5 \cos{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{u^{4}}{5}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = \frac{\int u^{4}\, du}{5}$
      
      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: $\frac{u^{5}}{25}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{5}{\left(5 x \right)}}{25}$
  The result is: $- \frac{\sin^{7}{\left(5 x \right)}}{35} + \frac{\sin^{5}{\left(5 x \right)}}{25}$
Method #3
1. Rewrite the integrand:
  
  $\left(1 - \sin^{2}{\left(5 x \right)}\right) \sin^{4}{\left(5 x \right)} \cos{\left(5 x \right)} = - \sin^{6}{\left(5 x \right)} \cos{\left(5 x \right)} + \sin^{4}{\left(5 x \right)} \cos{\left(5 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sin^{6}{\left(5 x \right)} \cos{\left(5 x \right)}\right)\, dx = - \int \sin^{6}{\left(5 x \right)} \cos{\left(5 x \right)}\, dx$
    1. Let $u = \sin{\left(5 x \right)}$ .
      
      Then let $du = 5 \cos{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$ :
      
      $\int \frac{u^{6}}{5}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{6}\, du = \frac{\int u^{6}\, du}{5}$
      
      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: $\frac{u^{7}}{35}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{7}{\left(5 x \right)}}{35}$
    So, the result is: $- \frac{\sin^{7}{\left(5 x \right)}}{35}$
  1. Let $u = \sin{\left(5 x \right)}$ .
    
    Then let $du = 5 \cos{\left(5 x \right)} dx$ and substitute $\frac{du}{5}$ :
    
    $\int \frac{u^{4}}{5}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = \frac{\int u^{4}\, du}{5}$
      
      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: $\frac{u^{5}}{25}$
    Now substitute $u$ back in:
    
    $\frac{\sin^{5}{\left(5 x \right)}}{25}$
  The result is: $- \frac{\sin^{7}{\left(5 x \right)}}{35} + \frac{\sin^{5}{\left(5 x \right)}}{25}$
Add the constant of integration:

$- \frac{\sin^{7}{\left(5 x \right)}}{35} + \frac{\sin^{5}{\left(5 x \right)}}{25}+ \mathrm{constant}$

The answer is:

$- \frac{\sin^{7}{\left(5 x \right)}}{35} + \frac{\sin^{5}{\left(5 x \right)}}{25}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                  
 |                                 7           5     
 |    3         4               sin (5*x)   sin (5*x)
 | cos (5*x)*sin (5*x) dx = C - --------- + ---------
 |                                  35          25   
/

$$\int \sin^{4}{\left(5 x \right)} \cos^{3}{\left(5 x \right)}\, dx = C - \frac{\sin^{7}{\left(5 x \right)}}{35} + \frac{\sin^{5}{\left(5 x \right)}}{25}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

     7         5   
  sin (5)   sin (5)
- ------- + -------
     35        25

$$\frac{\sin^{5}{\left(5 \right)}}{25} - \frac{\sin^{7}{\left(5 \right)}}{35}$$

     7         5   
  sin (5)   sin (5)
- ------- + -------
     35        25

$$\frac{\sin^{5}{\left(5 \right)}}{25} - \frac{\sin^{7}{\left(5 \right)}}{35}$$

-sin(5)^7/35 + sin(5)^5/25

Numerical answer [src]

-0.0111304977091004

-0.0111304977091004

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

Other calculators

How to use it?

Identical expressions

Integral of cos^3(5x)sin^4(5*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Integral of cos^3(5*x)*sin^4(5*x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of cos^3(5x)sin^4(5*x) dx