What is the integral from 0 to 2pi of sin^2(15x) ?

The integral from 0 to 2pi of sin^2(15x) is pi

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integral from 0 to 2pi of sin^2(15x)

\int_{0}^{2 π} sin^{2} (15 x) d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{0}^{2 π} sin^{2} (15 x) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} (1 - cos (30 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_{0}^{2 π} 1 - cos (30 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_{0}^{2 π} 1 d x - \int_{0}^{2 π} cos (30 x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (30 x) d x = 0

= \frac{1}{2} (2 π - 0)

Simplify

\frac{1}{2} (2 π - 0) : π

= π

x \to 4 lim (\frac{x ^{2} + 16 - 64}{x})

n = 0 \sum \infty ((\frac{1}{3})^{n})^{2}

\frac{d}{d x} (6 sin (2 π x))

\int 8 cos (θ) sin (θ) d θ

\frac{\partial}{\partial u} (3 cos (u))

What is the integral from 0 to 2pi of sin^2(15x) ?
The integral from 0 to 2pi of sin^2(15x) is pi