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Integral of (sinx-cosx)^2
Identical expressions
(sinx-cosx)^ two
( sinus of x minus co sinus of e of x) squared
( sinus of x minus co sinus of e of x) to the power of two
(sinx-cosx)2
sinx-cosx2
(sinx-cosx)²
(sinx-cosx) to the power of 2
sinx-cosx^2
(sinx-cosx)^2dx
Similar expressions
(sinx+cosx)^2

Solving integrals/ (sinx-cosx)^2

Integral of (sinx-cosx)^2 dx

The solution

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  1                      
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 |  (sin(x) - cos(x))  dx
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0

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

Integral((sin(x) - cos(x))^2, (x, 0, 1))

Detail solution

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

   2   
cos (1)

$$\cos^{2}{\left(1 \right)}$$

   2   
cos (1)

$$\cos^{2}{\left(1 \right)}$$

cos(1)^2

Numerical answer [src]

0.291926581726429

0.291926581726429

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 (sinx-cosx)^2 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2