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Integral of sinx/1-cos^2x dx

The solution

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

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

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

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\sin{\left(x \right)}}{1}\, dx = \int \sin{\left(x \right)}\, dx$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $- \cos{\left(x \right)}$
  So, the result is: $- \cos{\left(x \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \cos^{2}{\left(x \right)}\right)\, dx = - \int \cos^{2}{\left(x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{2}{\left(x \right)} = \frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(2 x \right)}}{2}\, dx = \frac{\int \cos{\left(2 x \right)}\, dx}{2}$
      1. 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: $\frac{\sin{\left(2 x \right)}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
    The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
  So, the result is: $- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}$
The result is: $- \frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4} - \cos{\left(x \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

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

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

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

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

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

Numerical answer [src]

-0.26762666257456

-0.26762666257456

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

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Identical expressions

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Integral of sinx/1-cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution