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Identical expressions
(nine *x^ two + nine *x+ eleven)*cos3xdx
(9 multiply by x squared plus 9 multiply by x plus 11) multiply by co sinus of e of 3xdx
(nine multiply by x to the power of two plus nine multiply by x plus eleven) multiply by co sinus of e of 3xdx
(9*x2+9*x+11)*cos3xdx
9*x2+9*x+11*cos3xdx
(9*x²+9*x+11)*cos3xdx
(9*x to the power of 2+9*x+11)*cos3xdx
(9x^2+9x+11)cos3xdx
(9x2+9x+11)cos3xdx
9x2+9x+11cos3xdx
9x^2+9x+11cos3xdx
Similar expressions
(9*x^2-9*x+11)*cos3xdx
(9*x^2+9*x-11)*cos3xdx

Solving integrals/ (9*x^2+9*x+11)*cos3xdx

Integral of (9x^2+9x+11)*cos3xdx dx

The solution

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 pi                              
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 |  \9*x  + 9*x + 11/*cos(3*x) dx
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0

$$\int\limits_{0}^{\pi} \left(\left(9 x^{2} + 9 x\right) + 11\right) \cos{\left(3 x \right)}\, dx$$

Integral((9*x^2 + 9*x + 11)*cos(3*x), (x, 0, pi))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 3 x$ .
  
  Then let $du = 3 dx$ and substitute $du$ :
  
  $\int \left(\frac{u^{2} \cos{\left(u \right)}}{3} + u \cos{\left(u \right)} + \frac{11 \cos{\left(u \right)}}{3}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2} \cos{\left(u \right)}}{3}\, du = \frac{\int u^{2} \cos{\left(u \right)}\, du}{3}$
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = \cos{\left(u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      Now evaluate the sub-integral.
      
      Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = 2 u$ and let $\operatorname{dv}{\left(u \right)} = \sin{\left(u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 2 \cos{\left(u \right)}\right)\, du = - 2 \int \cos{\left(u \right)}\, du$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $- 2 \sin{\left(u \right)}$
      
      So, the result is: $\frac{u^{2} \sin{\left(u \right)}}{3} + \frac{2 u \cos{\left(u \right)}}{3} - \frac{2 \sin{\left(u \right)}}{3}$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = \cos{\left(u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      Now evaluate the sub-integral.
    2. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{11 \cos{\left(u \right)}}{3}\, du = \frac{11 \int \cos{\left(u \right)}\, du}{3}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{11 \sin{\left(u \right)}}{3}$
    The result is: $\frac{u^{2} \sin{\left(u \right)}}{3} + u \sin{\left(u \right)} + \frac{2 u \cos{\left(u \right)}}{3} + 3 \sin{\left(u \right)} + \cos{\left(u \right)}$
  Now substitute $u$ back in:
  
  $3 x^{2} \sin{\left(3 x \right)} + 3 x \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} + 3 \sin{\left(3 x \right)} + \cos{\left(3 x \right)}$
Method #2
1. Rewrite the integrand:
  
  $\left(\left(9 x^{2} + 9 x\right) + 11\right) \cos{\left(3 x \right)} = 9 x^{2} \cos{\left(3 x \right)} + 9 x \cos{\left(3 x \right)} + 11 \cos{\left(3 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 9 x^{2} \cos{\left(3 x \right)}\, dx = 9 \int x^{2} \cos{\left(3 x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin{\left(3 x \right)}}{27}$
    So, the result is: $3 x^{2} \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} - \frac{2 \sin{\left(3 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 9 x \cos{\left(3 x \right)}\, dx = 9 \int x \cos{\left(3 x \right)}\, dx$
    1. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{\cos{\left(3 x \right)}}{9}$
    So, the result is: $3 x \sin{\left(3 x \right)} + \cos{\left(3 x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 11 \cos{\left(3 x \right)}\, dx = 11 \int \cos{\left(3 x \right)}\, dx$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $\frac{11 \sin{\left(3 x \right)}}{3}$
  The result is: $3 x^{2} \sin{\left(3 x \right)} + 3 x \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} + 3 \sin{\left(3 x \right)} + \cos{\left(3 x \right)}$
Method #3
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 9 x^{2} + 9 x + 11$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 18 x + 9$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    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:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
    
    $\frac{\sin{\left(3 x \right)}}{3}$
  Now evaluate the sub-integral.
2. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = 6 x + 3$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 6$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 3 x$ .
    
    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:
      
      $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{3}$
      
      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:
    
    $- \frac{\cos{\left(3 x \right)}}{3}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- 2 \cos{\left(3 x \right)}\right)\, dx = - 2 \int \cos{\left(3 x \right)}\, dx$
  1. Let $u = 3 x$ .
    
    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:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
    
    $\frac{\sin{\left(3 x \right)}}{3}$
  So, the result is: $- \frac{2 \sin{\left(3 x \right)}}{3}$
Method #4
1. Rewrite the integrand:
  
  $\left(\left(9 x^{2} + 9 x\right) + 11\right) \cos{\left(3 x \right)} = 9 x^{2} \cos{\left(3 x \right)} + 9 x \cos{\left(3 x \right)} + 11 \cos{\left(3 x \right)}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 9 x^{2} \cos{\left(3 x \right)}\, dx = 9 \int x^{2} \cos{\left(3 x \right)}\, dx$
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 2 x$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(x \right)} = \frac{2 x}{3}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = \frac{2}{3}$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{2 \cos{\left(3 x \right)}}{9}\right)\, dx = - \frac{2 \int \cos{\left(3 x \right)}\, dx}{9}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{2 \sin{\left(3 x \right)}}{27}$
    So, the result is: $3 x^{2} \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} - \frac{2 \sin{\left(3 x \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 9 x \cos{\left(3 x \right)}\, dx = 9 \int x \cos{\left(3 x \right)}\, dx$
    1. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(3 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(3 x \right)}}{3}\, dx = \frac{\int \sin{\left(3 x \right)}\, dx}{3}$
      
      Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\sin{\left(u \right)}}{3}\, du$
      
      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:
      
      $- \frac{\cos{\left(3 x \right)}}{3}$
      
      So, the result is: $- \frac{\cos{\left(3 x \right)}}{9}$
    So, the result is: $3 x \sin{\left(3 x \right)} + \cos{\left(3 x \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 11 \cos{\left(3 x \right)}\, dx = 11 \int \cos{\left(3 x \right)}\, dx$
    1. Let $u = 3 x$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{\cos{\left(u \right)}}{3}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$
      
      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:
      
      $\frac{\sin{\left(3 x \right)}}{3}$
    So, the result is: $\frac{11 \sin{\left(3 x \right)}}{3}$
  The result is: $3 x^{2} \sin{\left(3 x \right)} + 3 x \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} + 3 \sin{\left(3 x \right)} + \cos{\left(3 x \right)}$
Add the constant of integration:

$3 x^{2} \sin{\left(3 x \right)} + 3 x \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} + 3 \sin{\left(3 x \right)} + \cos{\left(3 x \right)}+ \mathrm{constant}$

The answer is:

$3 x^{2} \sin{\left(3 x \right)} + 3 x \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} + 3 \sin{\left(3 x \right)} + \cos{\left(3 x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                       
 |                                                                                                        
 | /   2           \                                                                 2                    
 | \9*x  + 9*x + 11/*cos(3*x) dx = C + 3*sin(3*x) + 2*x*cos(3*x) + 3*x*sin(3*x) + 3*x *sin(3*x) + cos(3*x)
 |                                                                                                        
/

$$\int \left(\left(9 x^{2} + 9 x\right) + 11\right) \cos{\left(3 x \right)}\, dx = C + 3 x^{2} \sin{\left(3 x \right)} + 3 x \sin{\left(3 x \right)} + 2 x \cos{\left(3 x \right)} + 3 \sin{\left(3 x \right)} + \cos{\left(3 x \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-2 - 2*pi

$$- 2 \pi - 2$$

-2 - 2*pi

$$- 2 \pi - 2$$

-2 - 2*pi

Numerical answer [src]

-8.28318530717957

-8.28318530717957

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

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Integral of (9x^2+9x+11)*cos3xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

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Identical expressions

Similar expressions

Integral of (9*x^2+9*x+11)*cos3xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4

Integral of (9x^2+9x+11)*cos3xdx dx