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Identical expressions
sin(two x)/sqrt(one -4sin(x)^2)
sinus of (2x) divide by square root of (1 minus 4 sinus of (x) squared )
sinus of (two x) divide by square root of (one minus 4 sinus of (x) squared )
sin(2x)/√(1-4sin(x)^2)
sin(2x)/sqrt(1-4sin(x)2)
sin2x/sqrt1-4sinx2
sin(2x)/sqrt(1-4sin(x)²)
sin(2x)/sqrt(1-4sin(x) to the power of 2)
sin2x/sqrt1-4sinx^2
sin(2x) divide by sqrt(1-4sin(x)^2)
sin(2x)/sqrt(1-4sin(x)^2)dx
Similar expressions
sin(2x)/sqrt(1+4sin(x)^2)
sin(2x)/sqrt(1-4sinx^2)

Solving integrals/ sin(2x)/sqrt(1-4sin(x)^2)

Integral of sin(2x)/sqrt(1-4sin(x)^2) dx

The solution

You have entered [src]

  1                      
  /                      
 |                       
 |       sin(2*x)        
 |  ------------------ dx
 |     _______________   
 |    /          2       
 |  \/  1 - 4*sin (x)    
 |                       
/                        
0

$$\int\limits_{0}^{1} \frac{\sin{\left(2 x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}\, dx$$

Integral(sin(2*x)/(sqrt(1 - 4*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:
  
  $\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}\, dx$
  1. Let $u = 1 - 4 \sin^{2}{\left(x \right)}$ .
    
    Then let $du = - 8 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $- \frac{du}{8}$ :
    
    $\int \frac{1}{64 \sqrt{u}}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{8 \sqrt{u}}\right)\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{8}$
      
      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: $- \frac{\sqrt{u}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{4}$
  So, the result is: $- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{2}$
Method #2
1. Rewrite the integrand:
  
  $\frac{\sin{\left(2 x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}} = \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}\, dx$
  1. Let $u = 1 - 4 \sin^{2}{\left(x \right)}$ .
    
    Then let $du = - 8 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $- \frac{du}{8}$ :
    
    $\int \frac{1}{64 \sqrt{u}}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{8 \sqrt{u}}\right)\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{8}$
      
      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: $- \frac{\sqrt{u}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{4}$
  So, the result is: $- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{2}$
Now simplify:

$- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{2}$
Add the constant of integration:

$- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{2}+ \mathrm{constant}$

The answer is:

$- \frac{\sqrt{1 - 4 \sin^{2}{\left(x \right)}}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

       _______________
      /          2    
1   \/  1 - 4*sin (1) 
- - ------------------
2           2

$$\frac{1}{2} - \frac{\sqrt{1 - 4 \sin^{2}{\left(1 \right)}}}{2}$$

       _______________
      /          2    
1   \/  1 - 4*sin (1) 
- - ------------------
2           2

$$\frac{1}{2} - \frac{\sqrt{1 - 4 \sin^{2}{\left(1 \right)}}}{2}$$

Упростить

Numerical answer [src]

(0.430126213598387 - 0.783630810530614j)

(0.430126213598387 - 0.783630810530614j)

The graph

Integral of sin(2x)/sqrt(1-4sin(x)^2) dx

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of sin(2x)/sqrt(1-4sin(x)^2) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2