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

The integral from 0 to 2pi of cos^2(nx) is pi+1/(4n)sin(4pin)

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

\int_{0}^{2 π} cos^{2} (n x) d x

π + \frac{1}{4 n} sin (4 πn)

Solution steps

\int_{0}^{2 π} cos^{2} (n x) d x

Rewrite using trig identities

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

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 n x) d x = \frac{sin ( 4 πn )}{2 n}

= \frac{1}{2} (2 π + \frac{sin ( 4 πn )}{2 n})

Simplify

\frac{1}{2} (2 π + \frac{sin ( 4 πn )}{2 n}) : π + \frac{1}{4 n} sin (4 πn)

= π + \frac{1}{4 n} sin (4 πn)

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What is the integral from 0 to 2pi of cos^2(nx) ?
The integral from 0 to 2pi of cos^2(nx) is pi+1/(4n)sin(4pin)