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Integral of d{x}:
Integral of 1/(1+x^2)
Integral of 1/x²
Integral of -2/x
Integral of -x
Identical expressions
sin(3x)/sqrt(cos(3x))
sinus of (3x) divide by square root of ( co sinus of e of (3x))
sin(3x)/√(cos(3x))
sin3x/sqrtcos3x
sin(3x) divide by sqrt(cos(3x))
sin(3x)/sqrt(cos(3x))dx

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

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

The solution

You have entered [src]

  1                
  /                
 |                 
 |    sin(3*x)     
 |  ------------ dx
 |    __________   
 |  \/ cos(3*x)    
 |                 
/                  
0

$$\int\limits_{0}^{1} \frac{\sin{\left(3 x \right)}}{\sqrt{\cos{\left(3 x \right)}}}\, dx$$

Integral(sin(3*x)/sqrt(cos(3*x)), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\sin{\left(u \right)}}{3 \sqrt{\cos{\left(u \right)}}}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{\sin{\left(u \right)}}{\sqrt{\cos{\left(u \right)}}}\, du = \frac{\int \frac{\sin{\left(u \right)}}{\sqrt{\cos{\left(u \right)}}}\, du}{3}$
    1. Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$ :
      
      $\int \left(- \frac{1}{\sqrt{u}}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{\sqrt{u}}\, du = - \int \frac{1}{\sqrt{u}}\, du$
      
      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: $- 2 \sqrt{u}$
      
      Now substitute $u$ back in:
      
      $- 2 \sqrt{\cos{\left(u \right)}}$
    So, the result is: $- \frac{2 \sqrt{\cos{\left(u \right)}}}{3}$
  Now substitute $u$ back in:
  
  $- \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}$
Method #2
1. Let $u = \sqrt{\cos{\left(3 x \right)}}$ .
  
  Then let $du = - \frac{3 \sin{\left(3 x \right)} dx}{2 \sqrt{\cos{\left(3 x \right)}}}$ and substitute $- \frac{2 du}{3}$ :
  
  $\int \left(- \frac{2}{3}\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\text{False}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
    So, the result is: $- \frac{2 u}{3}$
  Now substitute $u$ back in:
  
  $- \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                    
 |                           __________
 |   sin(3*x)            2*\/ cos(3*x) 
 | ------------ dx = C - --------------
 |   __________                3       
 | \/ cos(3*x)                         
 |                                     
/

$$\int \frac{\sin{\left(3 x \right)}}{\sqrt{\cos{\left(3 x \right)}}}\, dx = C - \frac{2 \sqrt{\cos{\left(3 x \right)}}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

        ________
2   2*\/ cos(3) 
- - ------------
3        3

$$\frac{2}{3} - \frac{2 \sqrt{\cos{\left(3 \right)}}}{3}$$

        ________
2   2*\/ cos(3) 
- - ------------
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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Identical expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2