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Solving integrals/ sin2x/(1+sin^2x)

Integral of sin2x/(1+sin^2x) dx

The solution

You have entered [src]

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

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

Integral(sin(2*x)/(1 + sin(x)^2), (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:
  
  $\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{2}{\left(x \right)} + 1}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{2}{\left(x \right)} + 1}\, dx$
  1. Let $u = \sin^{2}{\left(x \right)} + 1$ .
    
    Then let $du = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(\sin^{2}{\left(x \right)} + 1 \right)}}{2}$
  So, the result is: $\log{\left(\sin^{2}{\left(x \right)} + 1 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\sin{\left(2 x \right)}}{\sin^{2}{\left(x \right)} + 1} = \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{2}{\left(x \right)} + 1}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{2}{\left(x \right)} + 1}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{2}{\left(x \right)} + 1}\, dx$
  1. Let $u = \sin^{2}{\left(x \right)} + 1$ .
    
    Then let $du = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $\frac{du}{2}$ :
    
    $\int \frac{1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{2 u}\, du = \frac{\int \frac{1}{u}\, du}{2}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{2}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(\sin^{2}{\left(x \right)} + 1 \right)}}{2}$
  So, the result is: $\log{\left(\sin^{2}{\left(x \right)} + 1 \right)}$
Add the constant of integration:

$\log{\left(\sin^{2}{\left(x \right)} + 1 \right)}+ \mathrm{constant}$

The answer is:

$\log{\left(\sin^{2}{\left(x \right)} + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

$$\log \left(\sin ^2x+1\right)$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   /       2   \
log\1 + sin (1)/

$$\log \left(\sin ^21+1\right)$$

   /       2   \
log\1 + sin (1)/

$$\log{\left(\sin^{2}{\left(1 \right)} + 1 \right)}$$

Упростить

Numerical answer [src]

0.53536607938024

0.53536607938024

The graph

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

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How to use it?

Identical expressions

Similar expressions

Integral of sin2x/(1+sin^2x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2