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Identical expressions
(x^ two -x+ one)*sin4x
(x squared minus x plus 1) multiply by sinus of 4x
(x to the power of two minus x plus one) multiply by sinus of 4x
(x2-x+1)*sin4x
x2-x+1*sin4x
(x²-x+1)*sin4x
(x to the power of 2-x+1)*sin4x
(x^2-x+1)sin4x
(x2-x+1)sin4x
x2-x+1sin4x
x^2-x+1sin4x
(x^2-x+1)*sin4xdx
Similar expressions
(x^2-x-1)*sin4x
(x^2+x+1)*sin4x

Solving integrals/ (x^2-x+1)*sin4x

Integral of (x^2-x+1)*sin4x dx

The solution

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0

$$\int\limits_{0}^{1} \left(x^{2} - x + 1\right) \sin{\left(4 x \right)}\, dx$$

Integral((x^2 - x + 1)*sin(4*x), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = - x$ .
  
  Then let $du = - dx$ and substitute $du$ :
  
  $\int \left(u^{2} \sin{\left(4 u \right)} + u \sin{\left(4 u \right)} + \sin{\left(4 u \right)}\right)\, du$
  1. Integrate term-by-term:
    1. Use integration by parts:
      
      $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
      
      Let $u{\left(u \right)} = u^{2}$ and let $\operatorname{dv}{\left(u \right)} = \sin{\left(4 u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 2 u$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 u \right)}}{4}$
      
      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 $u{\left(u \right)} = - \frac{u}{2}$ and let $\operatorname{dv}{\left(u \right)} = \cos{\left(4 u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = - \frac{1}{2}$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      Now evaluate the sub-integral.
    3. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\sin{\left(4 u \right)}}{8}\right)\, du = - \frac{\int \sin{\left(4 u \right)}\, du}{8}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 u \right)}}{4}$
      
      So, the result is: $\frac{\cos{\left(4 u \right)}}{32}$
    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 $\operatorname{dv}{\left(u \right)} = \sin{\left(4 u \right)}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 u \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 u \right)}}{4}\right)\, du = - \frac{\int \cos{\left(4 u \right)}\, du}{4}$
      
      Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 u \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 u \right)}}{16}$
    1. Let $u = 4 u$ .
      
      Then let $du = 4 du$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 u \right)}}{4}$
    The result is: $- \frac{u^{2} \cos{\left(4 u \right)}}{4} + \frac{u \sin{\left(4 u \right)}}{8} - \frac{u \cos{\left(4 u \right)}}{4} + \frac{\sin{\left(4 u \right)}}{16} - \frac{7 \cos{\left(4 u \right)}}{32}$
  Now substitute $u$ back in:
  
  $- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{7 \cos{\left(4 x \right)}}{32}$
Method #2
1. Rewrite the integrand:
  
  $\left(x^{2} - x + 1\right) \sin{\left(4 x \right)} = x^{2} \sin{\left(4 x \right)} - x \sin{\left(4 x \right)} + \sin{\left(4 x \right)}$
2. Integrate term-by-term:
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    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 $u{\left(x \right)} = - \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x \sin{\left(4 x \right)}\right)\, dx = - \int x \sin{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 x \right)}}{16}$
    So, the result is: $\frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  The result is: $- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{7 \cos{\left(4 x \right)}}{32}$
Method #3
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = x^{2} - x + 1$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = 2 x - 1$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  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 $u{\left(x \right)} = \frac{1}{4} - \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
  
  Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\cos{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(4 x \right)}}{4}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
Method #4
1. Rewrite the integrand:
  
  $\left(x^{2} - x + 1\right) \sin{\left(4 x \right)} = x^{2} \sin{\left(4 x \right)} - x \sin{\left(4 x \right)} + \sin{\left(4 x \right)}$
2. Integrate term-by-term:
  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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = 2 x$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    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 $u{\left(x \right)} = - \frac{x}{2}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(4 x \right)}$ .
    
    Then $\operatorname{du}{\left(x \right)} = - \frac{1}{2}$ .
    
    To find $v{\left(x \right)}$ :
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
    Now evaluate the sub-integral.
  3. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{\sin{\left(4 x \right)}}{8}\right)\, dx = - \frac{\int \sin{\left(4 x \right)}\, dx}{8}$
    1. Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
    So, the result is: $\frac{\cos{\left(4 x \right)}}{32}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- x \sin{\left(4 x \right)}\right)\, dx = - \int x \sin{\left(4 x \right)}\, dx$
    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 $\operatorname{dv}{\left(x \right)} = \sin{\left(4 x \right)}$ .
      
      Then $\operatorname{du}{\left(x \right)} = 1$ .
      
      To find $v{\left(x \right)}$ :
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\sin{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos{\left(4 x \right)}}{4}$
      
      Now evaluate the sub-integral.
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{\cos{\left(4 x \right)}}{4}\right)\, dx = - \frac{\int \cos{\left(4 x \right)}\, dx}{4}$
      
      Let $u = 4 x$ .
      
      Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
      
      $\int \frac{\cos{\left(u \right)}}{16}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{4}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$
      
      The integral of cosine is sine:
      
      $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      
      So, the result is: $\frac{\sin{\left(u \right)}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin{\left(4 x \right)}}{4}$
      
      So, the result is: $- \frac{\sin{\left(4 x \right)}}{16}$
    So, the result is: $\frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16}$
  1. Let $u = 4 x$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{\sin{\left(u \right)}}{16}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)}}{4}\, du = \frac{\int \sin{\left(u \right)}\, du}{4}$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $- \frac{\cos{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $- \frac{\cos{\left(4 x \right)}}{4}$
  The result is: $- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{7 \cos{\left(4 x \right)}}{32}$
Add the constant of integration:

$- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{7 \cos{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer is:

$- \frac{x^{2} \cos{\left(4 x \right)}}{4} + \frac{x \sin{\left(4 x \right)}}{8} + \frac{x \cos{\left(4 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{16} - \frac{7 \cos{\left(4 x \right)}}{32}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                            
 |                                                         2                                   
 | / 2        \                   7*cos(4*x)   sin(4*x)   x *cos(4*x)   x*cos(4*x)   x*sin(4*x)
 | \x  - x + 1/*sin(4*x) dx = C - ---------- - -------- - ----------- + ---------- + ----------
 |                                    32          16           4            4            8     
/

$${{{{8\,x\,\sin \left(4\,x\right)+\left(2-16\,x^2\right)\,\cos \left(4\,x\right)}\over{16}}-{{\sin \left(4\,x\right)-4\,x\,\cos \left(4\,x\right)}\over{4}}-\cos \left(4\,x\right)}\over{4}}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

7    7*cos(4)   sin(4)
-- - -------- + ------
32      32        16

$${{2\,\sin 4-7\,\cos 4}\over{32}}+{{7}\over{32}}$$

7    7*cos(4)   sin(4)
-- - -------- + ------
32      32        16

$$\frac{\sin{\left(4 \right)}}{16} - \frac{7 \cos{\left(4 \right)}}{32} + \frac{7}{32}$$

Simplify

Numerical answer [src]

0.31443438610717

0.31443438610717

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (x^2-x+1)*sin4x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Method #4