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

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

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

\int_{0}^{π} cos^{2} (n x) d x

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

Solution steps

\int_{0}^{π} cos^{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})

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