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Integral of 1-sin^2x/1-cos2x dx

The solution

You have entered [src]

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

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

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

Detail solution

Integrate term-by-term:
1. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\sin^{2}{\left(x \right)}}{1}\right)\, dx = - \int \sin^{2}{\left(x \right)}\, dx$
    1. Rewrite the integrand:
      
      $\sin^{2}{\left(x \right)} = \frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}$
    2. Integrate term-by-term:
      1. The integral of a constant is the constant times the variable of integration:
        
        $\int \frac{1}{2}\, dx = \frac{x}{2}$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
        
        Let $u = 2 x$ .
        
        Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
        
        So, the result is: $\frac{\sin{\left(u \right)}}{2}$
        
        Now substitute $u$ back in:
        
        $\frac{\sin{\left(2 x \right)}}{2}$
        
        So, the result is: $- \frac{\sin{\left(2 x \right)}}{4}$
      The result is: $\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$
    So, the result is: $- \frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \cos{\left(2 x \right)}\right)\, dx = - \int \cos{\left(2 x \right)}\, dx$
  1. Let $u = 2 x$ .
    
    Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{\cos{\left(u \right)}}{2}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
      1. The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      So, the result is: $\frac{\sin{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 x \right)}}{2}$
  So, the result is: $- \frac{\sin{\left(2 x \right)}}{2}$
The result is: $\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$
Add the constant of integration:

$\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}+ \mathrm{constant}$

The answer is:

$\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                              
 |                                               
 | /       2              \                      
 | |    sin (x)           |          x   sin(2*x)
 | |1 - ------- - cos(2*x)| dx = C + - - --------
 | \       1              /          2      4    
 |                                               
/

$$\int \left(\left(- \frac{\sin^{2}{\left(x \right)}}{1} + 1\right) - \cos{\left(2 x \right)}\right)\, dx = C + \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1   sin(2)   cos(1)*sin(1)
- - ------ + -------------
2     2            2

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

1   sin(2)   cos(1)*sin(1)
- - ------ + -------------
2     2            2

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

1/2 - sin(2)/2 + cos(1)*sin(1)/2

Numerical answer [src]

0.27267564329358

0.27267564329358

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

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Integral of 1-sin^2x/1-cos2x dx

Limits of integration:

The graph:

Piecewise:

The solution