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sin(2x)/(cosx)^ three
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sin(2x)/(cosx)^3dx

Solving integrals/ sin(2x)/(cosx)^3

Integral of sin(2x)/(cosx)^3 dx

The solution

You have entered [src]

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

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

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

Detail solution

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

$\frac{2}{\cos{\left(x \right)}}+ \mathrm{constant}$

The answer is:

$\frac{2}{\cos{\left(x \right)}}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                        
 |                         
 | sin(2*x)            2   
 | -------- dx = C + ------
 |    3              cos(x)
 | cos (x)                 
 |                         
/

$${{2}\over{\cos x}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       2   
-2 + ------
     cos(1)

$${{2}\over{\cos 1}}-2$$

       2   
-2 + ------
     cos(1)

$$-2 + \frac{2}{\cos{\left(1 \right)}}$$

Simplify

Numerical answer [src]

1.70163143536185

1.70163143536185

The graph

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

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Identical expressions

Integral of sin(2x)/(cosx)^3 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2