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

The integral of sin^2(x/6) is 1/2 (x-3sin(x/3))+C

integral of sin^2(x/6)

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

\frac{1}{2} (x - 3 sin (\frac{x}{3})) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} (1 - cos (\frac{x}{3})) 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 (\frac{x}{3}) 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 (\frac{x}{3}) d x)

\int 1 d x = x

\int cos (\frac{x}{3}) d x = 3 sin (\frac{x}{3})

= \frac{1}{2} (x - 3 sin (\frac{x}{3}))

Add a constant to the solution

= \frac{1}{2} (x - 3 sin (\frac{x}{3})) + C

\int 2 sec^{2} (2 x) tan (2 x) d x

\int \frac{7 e ^{7 x}}{x} d x

\int 2 x^{2} y cos (x) d y

\int \frac{x e ^{2 x}}{2 x + 1} d x

\int e^{2^{x}} d x

What is the integral of sin^2(x/6) ?
The integral of sin^2(x/6) is 1/2 (x-3sin(x/3))+C