Mister Exam

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Identical expressions
a*cos(two pix)+b*sin(2pix)*cos(2pix)+c*cos^2(2pix)
a multiply by co sinus of e of (2 Pi x) plus b multiply by sinus of (2 Pi x) multiply by co sinus of e of (2 Pi x) plus c multiply by co sinus of e of squared (2 Pi x)
a multiply by co sinus of e of (two Pi x) plus b multiply by sinus of (2 Pi x) multiply by co sinus of e of (2 Pi x) plus c multiply by co sinus of e of squared (2 Pi x)
a*cos(2pix)+b*sin(2pix)*cos(2pix)+c*cos2(2pix)
a*cos2pix+b*sin2pix*cos2pix+c*cos22pix
a*cos(2pix)+b*sin(2pix)*cos(2pix)+c*cos²(2pix)
a*cos(2pix)+b*sin(2pix)*cos(2pix)+c*cos to the power of 2(2pix)
acos(2pix)+bsin(2pix)cos(2pix)+ccos^2(2pix)
acos(2pix)+bsin(2pix)cos(2pix)+ccos2(2pix)
acos2pix+bsin2pixcos2pix+ccos22pix
acos2pix+bsin2pixcos2pix+ccos^22pix
a*cos(2pix)+b*sin(2pix)*cos(2pix)+c*cos^2(2pix)dx
Similar expressions
a*cos(2pix)-b*sin(2pix)*cos(2pix)+c*cos^2(2pix)
a*cos(2pix)+b*sin(2pix)*cos(2pix)-c*cos^2(2pix)

Solving integrals/ a*cos(2pix)+b*sin(2pix)*cos(2pix)+c*cos^2(2pix)

Integral of acos(2pix)+bsin(2pix)cos(2pix)+ccos^2(2pix) dx

The solution

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  1                                                                
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 |  /                                                 2        \   
 |  \a*cos(2*pi*x) + b*sin(2*pi*x)*cos(2*pi*x) + c*cos (2*pi*x)/ dx
 |                                                                 
/                                                                  
0

$$\int\limits_{0}^{1} \left(b \sin{\left(2 \pi x \right)} \cos{\left(2 \pi x \right)} + c \cos^{2}{\left(2 \pi x \right)} + a \cos{\left(2 \pi x \right)}\right)\, dx$$

Integral(a*cos(2*pi*x) + b*sin(2*pi*x)*cos(2*pi*x) + c*cos(2*pi*x)^2, (x, 0, 1))

Detail solution

Integrate term-by-term:
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int a \cos{\left(2 \pi x \right)}\, dx = a \int \cos{\left(2 \pi x \right)}\, dx$
  1. Let $u = 2 \pi x$ .
    
    Then let $du = 2 \pi dx$ and substitute $\frac{du}{2 \pi}$ :
    
    $\int \frac{\cos{\left(u \right)}}{4 \pi^{2}}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(u \right)}}{2 \pi}\, du = \frac{\int \cos{\left(u \right)}\, du}{2 \pi}$
      1. The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
      So, the result is: $\frac{\sin{\left(u \right)}}{2 \pi}$
    Now substitute $u$ back in:
    
    $\frac{\sin{\left(2 \pi x \right)}}{2 \pi}$
  So, the result is: $\frac{a \sin{\left(2 \pi x \right)}}{2 \pi}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int b \sin{\left(2 \pi x \right)} \cos{\left(2 \pi x \right)}\, dx = b \int \sin{\left(2 \pi x \right)} \cos{\left(2 \pi x \right)}\, dx$
  1. There are multiple ways to do this integral.
    Method #1
    1. Let $u = \cos{\left(2 \pi x \right)}$ .
      
      Then let $du = - 2 \pi \sin{\left(2 \pi x \right)} dx$ and substitute $- \frac{du}{2 \pi}$ :
      
      $\int \frac{u}{4 \pi^{2}}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u}{2 \pi}\right)\, du = - \frac{\int u\, du}{2 \pi}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{4 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(2 \pi x \right)}}{4 \pi}$
    Method #2
    1. Let $u = 2 \pi x$ .
      
      Then let $du = 2 \pi dx$ and substitute $\frac{du}{2 \pi}$ :
      
      $\int \frac{\sin{\left(u \right)} \cos{\left(u \right)}}{4 \pi^{2}}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(u \right)} \cos{\left(u \right)}}{2 \pi}\, du = \frac{\int \sin{\left(u \right)} \cos{\left(u \right)}\, du}{2 \pi}$
      
      Let $u = \cos{\left(u \right)}$ .
      
      Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$ :
      
      $\int u\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $- \frac{u^{2}}{2}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(u \right)}}{2}$
      
      So, the result is: $- \frac{\cos^{2}{\left(u \right)}}{4 \pi}$
      
      Now substitute $u$ back in:
      
      $- \frac{\cos^{2}{\left(2 \pi x \right)}}{4 \pi}$
    Method #3
    1. Let $u = \sin{\left(2 \pi x \right)}$ .
      
      Then let $du = 2 \pi \cos{\left(2 \pi x \right)} dx$ and substitute $\frac{du}{2 \pi}$ :
      
      $\int \frac{u}{4 \pi^{2}}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u}{2 \pi}\, du = \frac{\int u\, du}{2 \pi}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u\, du = \frac{u^{2}}{2}$
      
      So, the result is: $\frac{u^{2}}{4 \pi}$
      
      Now substitute $u$ back in:
      
      $\frac{\sin^{2}{\left(2 \pi x \right)}}{4 \pi}$
  So, the result is: $- \frac{b \cos^{2}{\left(2 \pi x \right)}}{4 \pi}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int c \cos^{2}{\left(2 \pi x \right)}\, dx = c \int \cos^{2}{\left(2 \pi x \right)}\, dx$
  1. Rewrite the integrand:
    
    $\cos^{2}{\left(2 \pi x \right)} = \frac{\cos{\left(4 \pi x \right)}}{2} + \frac{1}{2}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\cos{\left(4 \pi x \right)}}{2}\, dx = \frac{\int \cos{\left(4 \pi x \right)}\, dx}{2}$
      1. Let $u = 4 \pi x$ .
        
        Then let $du = 4 \pi dx$ and substitute $\frac{du}{4 \pi}$ :
        
        $\int \frac{\cos{\left(u \right)}}{16 \pi^{2}}\, du$
        
        The integral of a constant times a function is the constant times the integral of the function:
        
        $\int \frac{\cos{\left(u \right)}}{4 \pi}\, du = \frac{\int \cos{\left(u \right)}\, du}{4 \pi}$
        
        The integral of cosine is sine:
        
        $\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$
        
        So, the result is: $\frac{\sin{\left(u \right)}}{4 \pi}$
        
        Now substitute $u$ back in:
        
        $\frac{\sin{\left(4 \pi x \right)}}{4 \pi}$
      So, the result is: $\frac{\sin{\left(4 \pi x \right)}}{8 \pi}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{2}\, dx = \frac{x}{2}$
    The result is: $\frac{x}{2} + \frac{\sin{\left(4 \pi x \right)}}{8 \pi}$
  So, the result is: $c \left(\frac{x}{2} + \frac{\sin{\left(4 \pi x \right)}}{8 \pi}\right)$
The result is: $\frac{a \sin{\left(2 \pi x \right)}}{2 \pi} - \frac{b \cos^{2}{\left(2 \pi x \right)}}{4 \pi} + c \left(\frac{x}{2} + \frac{\sin{\left(4 \pi x \right)}}{8 \pi}\right)$
Now simplify:

$\frac{4 a \sin{\left(2 \pi x \right)} - 2 b \cos^{2}{\left(2 \pi x \right)} + c \left(4 \pi x + \sin{\left(4 \pi x \right)}\right)}{8 \pi}$
Add the constant of integration:

$\frac{4 a \sin{\left(2 \pi x \right)} - 2 b \cos^{2}{\left(2 \pi x \right)} + c \left(4 \pi x + \sin{\left(4 \pi x \right)}\right)}{8 \pi}+ \mathrm{constant}$

The answer is:

$\frac{4 a \sin{\left(2 \pi x \right)} - 2 b \cos^{2}{\left(2 \pi x \right)} + c \left(4 \pi x + \sin{\left(4 \pi x \right)}\right)}{8 \pi}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                                                          
 |                                                                                                                  2        
 | /                                                 2        \            /x   sin(4*pi*x)\   a*sin(2*pi*x)   b*cos (2*pi*x)
 | \a*cos(2*pi*x) + b*sin(2*pi*x)*cos(2*pi*x) + c*cos (2*pi*x)/ dx = C + c*|- + -----------| + ------------- - --------------
 |                                                                         \2       8*pi   /        2*pi            4*pi     
/

$${{c\,\left({{\sin \left(4\,\pi\,x\right)}\over{2}}+2\,\pi\,x\right) }\over{4\,\pi}}+{{a\,\sin \left(2\,\pi\,x\right)}\over{2\,\pi}}-{{b \,\cos ^2\left(2\,\pi\,x\right)}\over{4\,\pi}}$$

Simplify

The answer [src]

c
-
2

$${{c\,\sin \left(4\,\pi\right)+4\,a\,\sin \left(2\,\pi\right)-2\,b\, \cos ^2\left(2\,\pi\right)+4\,c\,\pi+2\,b}\over{8\,\pi}}$$

c
-
2

$$\frac{c}{2}$$

Simplify

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

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How to use it?

Identical expressions

Similar expressions

Integral of acos(2pix)+bsin(2pix)cos(2pix)+ccos^2(2pix) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of a*cos(2pix)+b*sin(2pix)*cos(2pix)+c*cos^2(2pix) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of acos(2pix)+bsin(2pix)cos(2pix)+ccos^2(2pix) dx