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

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

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

\int_{0}^{π} sin^{2} (9 x) d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{π} sin^{2} (9 x) d x

Rewrite using trig identities

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

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (18 x) d x = 0

= \frac{1}{2} (π - 0)

Simplify

\frac{1}{2} (π - 0) : \frac{π}{2}

= \frac{π}{2}

\frac{d}{d x} (- 2 x \cdot e^{- x^{2}})

(- 5, - 2), (5, 12)

y^{^{''}} + \frac{1}{2} y^{^{'}} + \frac{1}{16} y = 0, y (0) = 0, y^{^{'}} (0) = 5

\frac{\partial}{\partial y} (e^{x y} - z)

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