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

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

Popular >

integral of (sin(6x))^2

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

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

Solution steps

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

Simplify

= \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

y^{^{'}} = cos (t) y (1 - \frac{y}{100})

\int_{0}^{1} x^{2} e^{4 x} d x

n = 1 \sum \infty \frac{1}{4 ^{n} - 3 ^{n}}

2 (2 x y + 4 y - 3) d x + (x + 2)^{2} d y = 0

\int sin (5 x) cos (3 x) d x

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