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

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

The solution

You have entered [src]

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

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

$$2\,\sqrt{\sin ^2x+1}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

          _____________
         /        2    
-2 + 2*\/  1 + sin (1)

$$2\,\sqrt{\sin ^21+1}-2$$

          _____________
         /        2    
-2 + 2*\/  1 + sin (1)

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

Simplify

Numerical answer [src]

0.613865657047868

0.613865657047868

The graph

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2