Integral of cosxcos2xdx dx

The solution

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$$\int\limits_{0}^{1} \cos{\left(x \right)} \cos{\left(2 x \right)}\, dx$$

Integral(cos(x)*cos(2*x), (x, 0, 1))

Detail solution

Rewrite the integrand:

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

  cos(2)*sin(1)   2*cos(1)*sin(2)
- ------------- + ---------------
        3                3

$$- \frac{\sin{\left(1 \right)} \cos{\left(2 \right)}}{3} + \frac{2 \sin{\left(2 \right)} \cos{\left(1 \right)}}{3}$$

  cos(2)*sin(1)   2*cos(1)*sin(2)
- ------------- + ---------------
        3                3

$$- \frac{\sin{\left(1 \right)} \cos{\left(2 \right)}}{3} + \frac{2 \sin{\left(2 \right)} \cos{\left(1 \right)}}{3}$$

-cos(2)*sin(1)/3 + 2*cos(1)*sin(2)/3

Numerical answer [src]

0.444255493747259

0.444255493747259

The graph

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

Other calculators

How to use it?

Identical expressions

Integral of cosxcos2xdx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3