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Integral of cos2x/sqrt(3+4sin2x) dx

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The solution

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  1                      
  /                      
 |                       
 |       cos(2*x)        
 |  ------------------ dx
 |    ________________   
 |  \/ 3 + 4*sin(2*x)    
 |                       
/                        
0                        
01cos(2x)4sin(2x)+3dx\int\limits_{0}^{1} \frac{\cos{\left(2 x \right)}}{\sqrt{4 \sin{\left(2 x \right)} + 3}}\, dx
Integral(cos(2*x)/sqrt(3 + 4*sin(2*x)), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=2xu = 2 x.

      Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

      cos(u)24sin(u)+3du\int \frac{\cos{\left(u \right)}}{2 \sqrt{4 \sin{\left(u \right)} + 3}}\, du

      1. The integral of a constant times a function is the constant times the integral of the function:

        cos(u)4sin(u)+3du=cos(u)4sin(u)+3du2\int \frac{\cos{\left(u \right)}}{\sqrt{4 \sin{\left(u \right)} + 3}}\, du = \frac{\int \frac{\cos{\left(u \right)}}{\sqrt{4 \sin{\left(u \right)} + 3}}\, du}{2}

        1. Let u=4sin(u)+3u = 4 \sin{\left(u \right)} + 3.

          Then let du=4cos(u)dudu = 4 \cos{\left(u \right)} du and substitute du4\frac{du}{4}:

          14udu\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:

            1udu=1udu4\int \frac{1}{\sqrt{u}}\, du = \frac{\int \frac{1}{\sqrt{u}}\, du}{4}

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              1udu=2u\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}

            So, the result is: u2\frac{\sqrt{u}}{2}

          Now substitute uu back in:

          4sin(u)+32\frac{\sqrt{4 \sin{\left(u \right)} + 3}}{2}

        So, the result is: 4sin(u)+34\frac{\sqrt{4 \sin{\left(u \right)} + 3}}{4}

      Now substitute uu back in:

      4sin(2x)+34\frac{\sqrt{4 \sin{\left(2 x \right)} + 3}}{4}

    Method #2

    1. Let u=4sin(2x)+3u = \sqrt{4 \sin{\left(2 x \right)} + 3}.

      Then let du=4cos(2x)dx4sin(2x)+3du = \frac{4 \cos{\left(2 x \right)} dx}{\sqrt{4 \sin{\left(2 x \right)} + 3}} and substitute du4\frac{du}{4}:

      14du\int \frac{1}{4}\, du

      1. The integral of a constant times a function is the constant times the integral of the function:

        False\text{False}

        1. The integral of a constant is the constant times the variable of integration:

          1du=u\int 1\, du = u

        So, the result is: u4\frac{u}{4}

      Now substitute uu back in:

      4sin(2x)+34\frac{\sqrt{4 \sin{\left(2 x \right)} + 3}}{4}

  2. Add the constant of integration:

    4sin(2x)+34+constant\frac{\sqrt{4 \sin{\left(2 x \right)} + 3}}{4}+ \mathrm{constant}


The answer is:

4sin(2x)+34+constant\frac{\sqrt{4 \sin{\left(2 x \right)} + 3}}{4}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                              
 |                               ________________
 |      cos(2*x)               \/ 3 + 4*sin(2*x) 
 | ------------------ dx = C + ------------------
 |   ________________                  4         
 | \/ 3 + 4*sin(2*x)                             
 |                                               
/                                                
cos(2x)4sin(2x)+3dx=C+4sin(2x)+34\int \frac{\cos{\left(2 x \right)}}{\sqrt{4 \sin{\left(2 x \right)} + 3}}\, dx = C + \frac{\sqrt{4 \sin{\left(2 x \right)} + 3}}{4}
The graph
0.001.000.100.200.300.400.500.600.700.800.901.0-1.0
The answer [src]
    ___     ______________
  \/ 3    \/ 3 + 4*sin(2) 
- ----- + ----------------
    4            4        
34+3+4sin(2)4- \frac{\sqrt{3}}{4} + \frac{\sqrt{3 + 4 \sin{\left(2 \right)}}}{4}
=
=
    ___     ______________
  \/ 3    \/ 3 + 4*sin(2) 
- ----- + ----------------
    4            4        
34+3+4sin(2)4- \frac{\sqrt{3}}{4} + \frac{\sqrt{3 + 4 \sin{\left(2 \right)}}}{4}
-sqrt(3)/4 + sqrt(3 + 4*sin(2))/4
Numerical answer [src]
0.211055894381827
0.211055894381827

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