Pair of sines or cosines

From (1), we have

$$\int_0^\pi \cos mx \cos nx \d x = \int_0^\pi \sin mx \sin nx \d x$$

Substituting this into (2) gives

$$2 \int_0^\pi \cos mx \cos nx \d x = 2 \int_0^\pi \sin mx \sin nx \d x = 0$$

Hence, for $n\neq m$, we have

$$\int_0^\pi \cos mx \cos nx \d x = 0$$

and

$$\int_0^\pi \sin mx \sin nx \d x = 0$$

Alternatively, we can prove the same results by integrating by parts, as follows.

$$K := \int_0^\pi \cos mx \cos nx \d x$$

$$= \frac{1}{m} \biggl[ \sin mx \cos nx \biggr]_0^\pi + \frac{n}{m} \int_0^\pi \sin mx \sin nx \d x$$

$$= \frac{n}{m} K$$

(by (1)). Hence, $K=0$. This argument also holds only for $n\neq m$, since otherwise $nK/m$ becomes $K$ and so we get $K=K$, so that $K$ is not necessarily zero.

In the case when $n=m$, we have

$$\int_0^\pi \cos mx \cos nx \d x = \int_0^\pi \cos^2 nx \d x$$

$$= \frac{1}{2} \int_0^\pi \biggl( 1 + \cos 2nx \biggr) \d x$$

$$= \frac{1}{2} \biggl[ x + \frac{1}{2n} \sin 2nx \biggr]_0^\pi$$

$$= \frac{1}{2} \pi$$

Since (1) is valid for $n=m$ as well as $n\neq m$, we have

$$\int_0^\pi \cos mx \cos nx \d x = \int_0^\pi \sin mx \sin nx \d x = \frac{1}{2} \pi \delta_{mn}$$

where $\delta_{mn}$ is the Kronecker delta.