Integral of xsin(3x+1) dx

The solution

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

Integral(x*sin(3*x + 1), (x, 0, 1))

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x + 1 \right)}$ .

Then $\operatorname{du}{\left(x \right)} = 1$ .

To find $v{\left(x \right)}$ :
1. Let $u = 3 x + 1$ .
  
  Then let $du = 3 dx$ 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:
  
  $- \frac{\cos{\left(3 x + 1 \right)}}{3}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \left(- \frac{\cos{\left(3 x + 1 \right)}}{3}\right)\, dx = - \frac{\int \cos{\left(3 x + 1 \right)}\, dx}{3}$
1. Let $u = 3 x + 1$ .
  
  Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
  
  $\int \frac{\cos{\left(u \right)}}{3}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
    1. The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
    So, the result is: $\frac{\sin{\left(u \right)}}{3}$
  Now substitute $u$ back in:
  
  $\frac{\sin{\left(3 x + 1 \right)}}{3}$
So, the result is: $- \frac{\sin{\left(3 x + 1 \right)}}{9}$
Add the constant of integration:

$- \frac{x \cos{\left(3 x + 1 \right)}}{3} + \frac{\sin{\left(3 x + 1 \right)}}{9}+ \mathrm{constant}$

The answer is:

$- \frac{x \cos{\left(3 x + 1 \right)}}{3} + \frac{\sin{\left(3 x + 1 \right)}}{9}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                     
 |                         sin(1 + 3*x)   x*cos(1 + 3*x)
 | x*sin(3*x + 1) dx = C + ------------ - --------------
 |                              9               3       
/

$$\int x \sin{\left(3 x + 1 \right)}\, dx = C - \frac{x \cos{\left(3 x + 1 \right)}}{3} + \frac{\sin{\left(3 x + 1 \right)}}{9}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

  cos(4)   sin(1)   sin(4)
- ------ - ------ + ------
    3        9        9

$$- \frac{\sin{\left(1 \right)}}{9} + \frac{\sin{\left(4 \right)}}{9} - \frac{\cos{\left(4 \right)}}{3}$$

  cos(4)   sin(1)   sin(4)
- ------ - ------ + ------
    3        9        9

$$- \frac{\sin{\left(1 \right)}}{9} + \frac{\sin{\left(4 \right)}}{9} - \frac{\cos{\left(4 \right)}}{3}$$

-cos(4)/3 - sin(1)/9 + sin(4)/9

Numerical answer [src]

0.0402952647194457

0.0402952647194457

The graph

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

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Integral of xsin(3x+1) dx

Limits of integration:

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The solution