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

The integral from 0 to 2pi of cos(mx)cos(nx) is 1/2 ((sin(2pim+2pin))/(m+n)+(sin(2pim-2pin))/(m-n))

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

\int_{0}^{2 π} cos (m x) cos (n x) d x

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

Solution steps

\int_{0}^{2 π} cos (m x) cos (n x) d x

Rewrite using trig identities

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

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

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

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

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