What is the integral from 0 to 2pi of sin^2(9x) ?

The integral from 0 to 2pi of sin^2(9x) is pi

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integral from 0 to 2pi of sin^2(9x)

\int_{0}^{2 π} sin^{2} (9 x) d x

π

Decimal

3.14159 \dots

Solution steps

\int_{0}^{2 π} sin^{2} (9 x) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} (1 - cos (18 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 (18 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 (18 x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (18 x) d x = 0

= \frac{1}{2} (2 π - 0)

Simplify

\frac{1}{2} (2 π - 0) : π

= π

4 t

x \to 0 lim (ln (1 + sin (x)))

\frac{d y}{d x} = x^{7} y

\frac{d}{d x} (ln (sinh (x)))

\int cos (\frac{nπ x}{2}) d x

What is the integral from 0 to 2pi of sin^2(9x) ?
The integral from 0 to 2pi of sin^2(9x) is pi