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Identical expressions
sin(two +3ln(x))/x
sinus of (2 plus 3ln(x)) divide by x
sinus of (two plus 3ln(x)) divide by x
sin2+3lnx/x
sin(2+3ln(x)) divide by x
sin(2+3ln(x))/xdx
Similar expressions
sin(2-3ln(x))/x

Integral of sin(2+3ln(x))/x dx

The solution

You have entered [src]

  1                     
  /                     
 |                      
 |  sin(2 + 3*log(x))   
 |  ----------------- dx
 |          x           
 |                      
/                       
0

$$\int\limits_{0}^{1} \frac{\sin{\left(3 \log{\left(x \right)} + 2 \right)}}{x}\, dx$$

Integral(sin(2 + 3*log(x))/x, (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

  /                                            
 |                                             
 | sin(2 + 3*log(x))          cos(2 + 3*log(x))
 | ----------------- dx = C - -----------------
 |         x                          3        
 |                                             
/

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

Simplify

The answer [src]

   1   cos(2)  1   cos(2) 
<- - - ------, - - ------>
   3     3     3     3

$$\left\langle - \frac{1}{3} - \frac{\cos{\left(2 \right)}}{3}, \frac{1}{3} - \frac{\cos{\left(2 \right)}}{3}\right\rangle$$

   1   cos(2)  1   cos(2) 
<- - - ------, - - ------>
   3     3     3     3

$$\left\langle - \frac{1}{3} - \frac{\cos{\left(2 \right)}}{3}, \frac{1}{3} - \frac{\cos{\left(2 \right)}}{3}\right\rangle$$

AccumBounds(-1/3 - cos(2)/3, 1/3 - cos(2)/3)

Numerical answer [src]

0.0966483783510806

0.0966483783510806

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

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Identical expressions

Similar expressions

Integral of sin(2+3ln(x))/x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2