Integral of sinx*cos3x dx

The solution

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

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

Detail solution

Rewrite the integrand:

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

$\left(\sin^{2}{\left(x \right)} + \frac{1}{2}\right) \cos^{2}{\left(x \right)}$
Add the constant of integration:

$\left(\sin^{2}{\left(x \right)} + \frac{1}{2}\right) \cos^{2}{\left(x \right)}+ \mathrm{constant}$

The answer is:

$\left(\sin^{2}{\left(x \right)} + \frac{1}{2}\right) \cos^{2}{\left(x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                        2   
 |                             4      3*cos (x)
 | sin(x)*cos(3*x) dx = C - cos (x) + ---------
 |                                        2    
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

  1   cos(1)*cos(3)   3*sin(1)*sin(3)
- - + ------------- + ---------------
  8         8                8

$$-{{\cos 4-2\,\cos 2}\over{8}}-{{1}\over{8}}$$

  1   cos(1)*cos(3)   3*sin(1)*sin(3)
- - + ------------- + ---------------
  8         8                8

$$- \frac{1}{8} + \frac{\cos{\left(1 \right)} \cos{\left(3 \right)}}{8} + \frac{3 \sin{\left(1 \right)} \sin{\left(3 \right)}}{8}$$

Упростить

Numerical answer [src]

-0.147331256528834

-0.147331256528834

The graph

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

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Identical expressions

Similar expressions

Integral of sinx*cos3x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2