Integral of sin(z)*cos(z) dx

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  I                 
  /                 
 |                  
 |  sin(z)*cos(z) dz
 |                  
/                   
0

$$\int\limits_{0}^{i} \sin{\left(z \right)} \cos{\left(z \right)}\, dz$$

Integral(sin(z)*cos(z), (z, 0, i))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \sin{\left(z \right)}$ .
  
  Then let $du = \cos{\left(z \right)} dz$ and substitute $du$ :
  
  $\int u\, du$
  1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int u\, du = \frac{u^{2}}{2}$
  Now substitute $u$ back in:
  
  $\frac{\sin^{2}{\left(z \right)}}{2}$
Method #2
1. Let $u = \cos{\left(z \right)}$ .
  
  Then let $du = - \sin{\left(z \right)} dz$ and substitute $- du$ :
  
  $\int \left(- u\right)\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u\, du = - \int u\, du$
    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:
  
  $- \frac{\cos^{2}{\left(z \right)}}{2}$
Add the constant of integration:

$\frac{\sin^{2}{\left(z \right)}}{2}+ \mathrm{constant}$

The answer is:

$\frac{\sin^{2}{\left(z \right)}}{2}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                          2   
 |                        sin (z)
 | sin(z)*cos(z) dz = C + -------
 |                           2   
/

$$\int \sin{\left(z \right)} \cos{\left(z \right)}\, dz = C + \frac{\sin^{2}{\left(z \right)}}{2}$$

Simplify

The answer [src]

     2    
-sinh (1) 
----------
    2

$$- \frac{\sinh^{2}{\left(1 \right)}}{2}$$

     2    
-sinh (1) 
----------
    2

$$- \frac{\sinh^{2}{\left(1 \right)}}{2}$$

-sinh(1)^2/2

Numerical answer [src]

(-0.690548922770908 + 0.0j)

(-0.690548922770908 + 0.0j)

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

Method #1