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

The integral from 0 to pi of sin^2(mx) is pi-1/2 (pi+(sin(2pim))/(2m))

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

\int_{0}^{π} sin^{2} (m x) d x

π - \frac{1}{2} (π + \frac{sin ( 2 πm )}{2 m})

Solution steps

\int_{0}^{π} sin^{2} (m x) d x

Rewrite using trig identities

= \int_{0}^{π} 1 - cos^{2} (m 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

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

\int_{0}^{π} 1 d x = π

\int_{0}^{π} cos^{2} (m x) d x = \frac{1}{2} (π + \frac{sin ( 2 πm )}{2 m})

= π - \frac{1}{2} (π + \frac{sin ( 2 πm )}{2 m})

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What is the integral from 0 to pi of sin^2(mx) ?
The integral from 0 to pi of sin^2(mx) is pi-1/2 (pi+(sin(2pim))/(2m))