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

The integral from 0 to pi of (sin(nx))^2 is 1/2 (pi-(sin(2pin))/(2n))

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integral from 0 to pi of (sin(nx))^2

\int_{0}^{π} (sin (n x))^{2} d x

\frac{1}{2} (π - \frac{sin ( 2 πn )}{2 n})

Solution steps

\int_{0}^{π} (sin (n x))^{2} d x

Simplify

= \int_{0}^{π} sin^{2} (n x) d x

Rewrite using trig identities

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

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos (2 n x) d x = \frac{sin ( 2 πn )}{2 n}

= \frac{1}{2} (π - \frac{sin ( 2 πn )}{2 n})

What is the integral from 0 to pi of (sin(nx))^2 ?
The integral from 0 to pi of (sin(nx))^2 is 1/2 (pi-(sin(2pin))/(2n))