What is the integral of sin^2(6x) ?

The integral of sin^2(6x) is 1/2 (x-1/12 sin(12x))+C

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integral of sin^2(6x)

\int sin^{2} (6 x) d x

\frac{1}{2} (x - \frac{1}{12} sin (12 x)) + C

Solution steps

\int sin^{2} (6 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (12 x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int 1 - cos (12 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d x - \int cos (12 x) d x)

\int 1 d x = x

\int cos (12 x) d x = \frac{1}{12} sin (12 x)

= \frac{1}{2} (x - \frac{1}{12} sin (12 x))

Add a constant to the solution

= \frac{1}{2} (x - \frac{1}{12} sin (12 x)) + C

\frac{\partial}{\partial y} (y cos (yz))

\frac{d}{d x} (x^{3} + 3 x^{2} - 2)

y^{^{'}} = (1 + x) (1 + y)

x \to 0 lim (\frac{( 5 e ^{x} - 5 )}{x})

\frac{\partial}{\partial y} (x^{2} - 3 y^{2} + 7)

What is the integral of sin^2(6x) ?
The integral of sin^2(6x) is 1/2 (x-1/12 sin(12x))+C