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Solving integrals/ sin(3x)/sqrt(cos(3x))

Integral of sin(3x)/sqrt(cos(3x)) dx

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The solution

You have entered [src] 
  1                
  /                
 |                 
 |    sin(3*x)     
 |  ------------ dx
 |    __________   
 |  \/ cos(3*x)    
 |                 
/                  
0
∫01sin⁡(3x)cos⁡(3x) dx\int\limits_{0}^{1} \frac{\sin{\left(3 x \right)}}{\sqrt{\cos{\left(3 x \right)}}}\, dx 
Integral(sin(3*x)/sqrt(cos(3*x)), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=3xu = 3 x.

      Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

      ∫sin⁡(u)3cos⁡(u) du\int \frac{\sin{\left(u \right)}}{3 \sqrt{\cos{\left(u \right)}}}\, du

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

        ∫sin⁡(u)cos⁡(u) du=∫sin⁡(u)cos⁡(u) du3\int \frac{\sin{\left(u \right)}}{\sqrt{\cos{\left(u \right)}}}\, du = \frac{\int \frac{\sin{\left(u \right)}}{\sqrt{\cos{\left(u \right)}}}\, du}{3}

        1. Let u=cos⁡(u)u = \cos{\left(u \right)}.

          Then let du=−sin⁡(u)dudu = - \sin{\left(u \right)} du and substitute −du- du:

          ∫(−1u) du\int \left(- \frac{1}{\sqrt{u}}\right)\, du

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

            ∫1u du=−∫1u du\int \frac{1}{\sqrt{u}}\, du = - \int \frac{1}{\sqrt{u}}\, du

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

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

            So, the result is: −2u- 2 \sqrt{u}

          Now substitute uu back in:

          −2cos⁡(u)- 2 \sqrt{\cos{\left(u \right)}}

        So, the result is: −2cos⁡(u)3- \frac{2 \sqrt{\cos{\left(u \right)}}}{3}

      Now substitute uu back in:

      −2cos⁡(3x)3- \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}

    Method #2

    1. Let u=cos⁡(3x)u = \sqrt{\cos{\left(3 x \right)}}.

      Then let du=−3sin⁡(3x)dx2cos⁡(3x)du = - \frac{3 \sin{\left(3 x \right)} dx}{2 \sqrt{\cos{\left(3 x \right)}}} and substitute −2du3- \frac{2 du}{3}:

      ∫(−23) du\int \left(- \frac{2}{3}\right)\, 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:

          ∫1 du=u\int 1\, du = u

        So, the result is: −2u3- \frac{2 u}{3}

      Now substitute uu back in:

      −2cos⁡(3x)3- \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}

  2. Add the constant of integration:

    −2cos⁡(3x)3+constant- \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}+ \mathrm{constant}

The answer is:

−2cos⁡(3x)3+constant- \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                    
 |                           __________
 |   sin(3*x)            2*\/ cos(3*x) 
 | ------------ dx = C - --------------
 |   __________                3       
 | \/ cos(3*x)                         
 |                                     
/
∫sin⁡(3x)cos⁡(3x) dx=C−2cos⁡(3x)3\int \frac{\sin{\left(3 x \right)}}{\sqrt{\cos{\left(3 x \right)}}}\, dx = C - \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}
Simplify
The graph
0.000.050.100.150.200.250.300.350.400.450.50200-100
Plot the graph f(x)
The answer [src] 
        ________
2   2*\/ cos(3) 
- - ------------
3        3
23−2cos⁡(3)3\frac{2}{3} - \frac{2 \sqrt{\cos{\left(3 \right)}}}{3} 
=
= 
        ________
2   2*\/ cos(3) 
- - ------------
3        3
23−2cos⁡(3)3\frac{2}{3} - \frac{2 \sqrt{\cos{\left(3 \right)}}}{3} 
2/3 - 2*sqrt(cos(3))/3
Numerical answer [src] 
(0.579862216335156 - 0.795858425143173j)
(0.579862216335156 - 0.795858425143173j)

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

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