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Solving integrals/ sin3x/cos3x^-4

Integral of sin3x/cos3x^-4 dx

The solution

You have entered [src]

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

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

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

Detail solution

Let $u = 3 x$ .

Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :

$\int \frac{\sin{\left(u \right)} \cos^{4}{\left(u \right)}}{3}\, du$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \sin{\left(u \right)} \cos^{4}{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)} \cos^{4}{\left(u \right)}\, du}{3}$
  1. Let $u = \cos{\left(u \right)}$ .
    
    Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$ :
    
    $\int \left(- u^{4}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int u^{4}\, du = - \int u^{4}\, du$
      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:
    
    $- \frac{\cos^{5}{\left(u \right)}}{5}$
  So, the result is: $- \frac{\cos^{5}{\left(u \right)}}{15}$
Now substitute $u$ back in:

$- \frac{\cos^{5}{\left(3 x \right)}}{15}$
Add the constant of integration:

$- \frac{\cos^{5}{\left(3 x \right)}}{15}+ \mathrm{constant}$

The answer is:

$- \frac{\cos^{5}{\left(3 x \right)}}{15}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                              
 |                         5     
 |   sin(3*x)           cos (3*x)
 | ----------- dx = C - ---------
 | /    1    \              15   
 | |---------|                   
 | |   4     |                   
 | \cos (3*x)/                   
 |                               
/

$$\int \frac{\sin{\left(3 x \right)}}{\frac{1}{\cos^{4}{\left(3 x \right)}}}\, dx = C - \frac{\cos^{5}{\left(3 x \right)}}{15}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

        5   
1    cos (3)
-- - -------
15      15

$$\frac{1}{15} - \frac{\cos^{5}{\left(3 \right)}}{15}$$

        5   
1    cos (3)
-- - -------
15      15

$$\frac{1}{15} - \frac{\cos^{5}{\left(3 \right)}}{15}$$

1/15 - cos(3)^5/15

Numerical answer [src]

0.130063600784528

0.130063600784528

The graph

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

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Identical expressions

Similar expressions

Integral of sin3x/cos3x^-4 dx

Limits of integration:

The graph:

Piecewise:

The solution