Integral of 6sin6xdx dx

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$$\int\limits_{\frac{p}{2}}^{0} 6 \sin{\left(6 x \right)}\, dx$$

Integral(6*sin(6*x), (x, p/2, 0))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int 6 \sin{\left(6 x \right)}\, dx = 6 \int \sin{\left(6 x \right)}\, dx$
1. Let $u = 6 x$ .
  
  Then let $du = 6 dx$ and substitute $\frac{du}{6}$ :
  
  $\int \frac{\sin{\left(u \right)}}{6}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{6}$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    So, the result is: $- \frac{\cos{\left(u \right)}}{6}$
  Now substitute $u$ back in:
  
  $- \frac{\cos{\left(6 x \right)}}{6}$
So, the result is: $- \cos{\left(6 x \right)}$
Add the constant of integration:

$- \cos{\left(6 x \right)}+ \mathrm{constant}$

The answer is:

$- \cos{\left(6 x \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

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 | 6*sin(6*x) dx = C - cos(6*x)
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$$\int 6 \sin{\left(6 x \right)}\, dx = C - \cos{\left(6 x \right)}$$

Simplify

The answer [src]

-1 + cos(3*p)

$$\cos{\left(3 p \right)} - 1$$

-1 + cos(3*p)

$$\cos{\left(3 p \right)} - 1$$

-1 + cos(3*p)

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