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Similar expressions
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Solving integrals/ sin^3x/1+cos^2x

Integral of sin^3x/1+cos^2x dx

The solution

You have entered [src]

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

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

Simplify

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{\sin^{3}{\left(x \right)}}{1}\, dx = \int \sin^{3}{\left(x \right)}\, dx$
  1. Don't know the steps in finding this integral.
    
    But the integral is
    
    $\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
  So, the result is: $\frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
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 $\frac{du}{2}$ :
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du$
      1. The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}$
        
        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: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4}$
The result is: $\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}$
Add the constant of integration:

$\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer is:

$\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} + \frac{\cos^{3}{\left(x \right)}}{3} - \cos{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                            
 |                                                             
 | /   3             \                          3              
 | |sin (x)      2   |          x            cos (x)   sin(2*x)
 | |------- + cos (x)| dx = C + - - cos(x) + ------- + --------
 | \   1             /          2               3         4    
 |                                                             
/

$${{{{\sin \left(2\,x\right)}\over{2}}+x}\over{2}}+{{\cos ^3x}\over{3 }}-\cos x$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

                3                   
7            cos (1)   cos(1)*sin(1)
- - cos(1) + ------- + -------------
6               3            2

$${{3\,\sin 2+4\,\cos ^31-12\,\cos 1+14}\over{12}}$$

                3                   
7            cos (1)   cos(1)*sin(1)
- - cos(1) + ------- + -------------
6               3            2

$$- \cos{\left(1 \right)} + \frac{\cos^{3}{\left(1 \right)}}{3} + \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{2} + \frac{7}{6}$$

Simplify

Numerical answer [src]

0.906264919255279

0.906264919255279

The graph

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

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Identical expressions

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Integral of sin^3x/1+cos^2x dx

Limits of integration:

The graph:

Piecewise:

The solution